QBManager

Vector Algebra

87703 If \(3 \hat{i}-5 \hat{j}+2 \hat{k}, 7 \hat{i}+2 \hat{j}-4 \hat{k}, \hat{i}-3 \hat{j}+4 \hat{k}\) and \(-7 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}\) are position vectors of the points \(A, B, C\) and \(D\) respectively, then the angle between \(A B\) and \(C D\) is

1 0

2 \(\frac{\pi}{4}\)

3 \(\frac{\pi}{2}\)

4 \(\pi\)

TS EAMCET-2022-18.07.2022

Vector Algebra

87704 If \(a, b, c\) are distinct real numbers and \(P, Q, R\) are three points whose position vectors are respectively \(\quad \mathbf{a} \hat{i}+b \hat{j}+c \hat{k}, b \hat{i}+c \hat{j}+a \hat{k} \quad\) and \(\hat{\mathbf{i}}+\mathbf{a} \hat{\mathbf{j}}+\mathbf{b} \hat{\mathbf{k}}\), then \(\angle \mathrm{QPR}=\)

1 \(\cos ^{-1}(a+b+c)\)

2 \(\frac{\pi}{2}\)

3 \(\frac{\pi}{3}\)

4 \(\cos ^{-1}\left(\frac{a^2+b^2+c^2}{a b c}\right)\)

TS EAMCET-2018-05.05.2018

Vector Algebra

87705 \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) are non-coplanar vectors. If the position vector of the point of intersection of the line \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\mathbf{2} \overrightarrow{\mathrm{b}}+\mathbf{p}(\overrightarrow{\mathbf{a}}-\mathbf{2} \mathbf{c})\) and the plane \(\overrightarrow{\mathbf{r}}=3 \overrightarrow{\mathbf{a}}-\mathrm{q}(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathrm{b}})+K(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathbf{c}})\) is \(\overrightarrow{\mathbf{r}}=\mathbf{x} \overrightarrow{\mathbf{a}}+\mathbf{y} \vec{b}+z \overrightarrow{\mathbf{c}}\), then \(x\) y \(z\)

1 -8

2 8

3 12

4 -12

TS EAMCET-2021-05.08.2021

Vector Algebra

87706 If \(\vec{a}, \vec{b}, \vec{c}\) are the position vectors of the points A, B, C respectively, then match the items of list-I with those of list-II.
| List-I |List-II|
|
|A. \(\vec{a} =2 \hat{i}+3 \hat{j}+4 \hat{k}\),\ltbr>\(\vec{b} =3 \hat{i}+4 \hat{j}+2 \hat{k}\),\ltbr>\(\vec{c} =4 \hat{i}+2 \hat{j}+3 \hat{k}\)|I. A, B, C are collinear|
|B. \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}}=-3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\)|II. \(\triangle \mathrm{ABC}\) is an isosceles triangle|
|C. \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}}=-3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}\)|III. \(\triangle \mathrm{ABC}\) is a right angled triangle|
|D. \(\overrightarrow{\mathbf{a}} =\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}} =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}} =2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)|IV. \(\triangle \text { ABC}\) is a right-angled triangle
| | V. \(\Delta \text { ABC is }\),equilateral, triangle|
The correct match is

1 A-I, B-IV, C-III, D-II

2 A-I, B-II, C-III, D-IV

3 A-V, B-I, C-IV, D-II

4 A-V, B-I, C-III, D-II

Explanation:

(D) : Given position vectors are-
\(\overrightarrow{\mathrm{OA}}=\mathrm{a}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \Rightarrow \overrightarrow{\mathrm{OC}}=\mathrm{c}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{AB}}|=\sqrt{1^2+1^2+2^2}=\sqrt{6}\)
\(|\overrightarrow{\mathrm{BC}}|=\sqrt{1^2+2^2+1^2}=\sqrt{6}\)
And, \(|\overrightarrow{\mathrm{CA}}|=\sqrt{2^2+1^2+1^2}=\sqrt{6}\)
\(\therefore \triangle \mathrm{ABC}\) is an equilateral triangle. \(\mathrm{B}\), Given position vectors are
\(\overrightarrow{\mathrm{OA}}=\mathrm{a}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OC}}=\mathrm{c}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\)
\(\mid \overrightarrow{\mathrm{AB}}=\sqrt{4+4+16}=\sqrt{24}=2 \sqrt{6}\)
\(\mid \overrightarrow{\mathrm{BC}}=\sqrt{36+36+144}=216=6 \sqrt{6}\)
And,
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{16+16+64}=\sqrt{96}=4 \sqrt{6}\)
\(\because|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{CA}}|=|\overrightarrow{\mathrm{BC}}| \Rightarrow \mathrm{A}, \mathrm{B}, \mathrm{C}\) are collinear points.
C. given position vectors are
\(\overrightarrow{\mathrm{OA}}=a=2 \hat{i}-\hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{OB}}=b=\hat{i}-3 \hat{j}-5 \hat{k}\)
And,
\(\overrightarrow{\mathrm{OC}}=\mathrm{c}=-3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\)
\(\mid \overrightarrow{\mathrm{AB}}=\sqrt{1+4+36}=\sqrt{41}\)
\(\mid \overrightarrow{\mathrm{BC}}=\sqrt{16+1+1}=\sqrt{18}\)
And,
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{25+9+25}=\sqrt{59}\)
\(\because|\overrightarrow{\mathrm{CA}}|^2=|\overrightarrow{\mathrm{AB}}|^2+|\overrightarrow{\mathrm{BC}}|^2=\Delta \mathrm{ABC}\) is right angled triangle.
D. Given position sectors, \(\overrightarrow{O A}=a=\hat{i}+\hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{OC}}=\mathrm{c}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\therefore|\overrightarrow{\mathrm{AB}}|=\sqrt{0+1+4}=\sqrt{5}, \mid \overrightarrow{\mathrm{BC}}=\sqrt{1+9+4}=\sqrt{14}\)
And, \(|\overrightarrow{\mathrm{AC}}|=\sqrt{1+4+0}=\sqrt{5}\)
\(\because|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{BC}}| \text { and }|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{AC}}|>|\overrightarrow{\mathrm{BC}}|\)
So, \(\triangle A B C\) is isosceles triangle.

TS EAMCET-2020-11.09.2020

Vector Algebra

87707 If \(\vec{a}, \vec{b}, \vec{c}\) are three independent vectors and there exists a non zero scalar traid \((l, \mathrm{~m}, \mathrm{n})\) such that \(l(3 a+2 b+c)+\mathbf{m}(2 a+2 b+3 c)+\mathbf{n}(\mathbf{a}\) \(+2 b+5 c)=0\), then

1 \(l=\mathrm{m}=\mathrm{n}\)

2 \(l=\mathrm{n}\)

3 \(l=\mathrm{n}, \mathrm{m}+2 \mathrm{n}=0\)

4 \(\mathrm{m}+2 \mathrm{n}=0, l+\mathrm{n}=0\)

TS EAMCET-2020-14.09.2020

Vector Algebra

87703 If \(3 \hat{i}-5 \hat{j}+2 \hat{k}, 7 \hat{i}+2 \hat{j}-4 \hat{k}, \hat{i}-3 \hat{j}+4 \hat{k}\) and \(-7 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}\) are position vectors of the points \(A, B, C\) and \(D\) respectively, then the angle between \(A B\) and \(C D\) is

1 0

2 \(\frac{\pi}{4}\)

3 \(\frac{\pi}{2}\)

4 \(\pi\)

TS EAMCET-2022-18.07.2022

Vector Algebra

87704 If \(a, b, c\) are distinct real numbers and \(P, Q, R\) are three points whose position vectors are respectively \(\quad \mathbf{a} \hat{i}+b \hat{j}+c \hat{k}, b \hat{i}+c \hat{j}+a \hat{k} \quad\) and \(\hat{\mathbf{i}}+\mathbf{a} \hat{\mathbf{j}}+\mathbf{b} \hat{\mathbf{k}}\), then \(\angle \mathrm{QPR}=\)

1 \(\cos ^{-1}(a+b+c)\)

2 \(\frac{\pi}{2}\)

3 \(\frac{\pi}{3}\)

4 \(\cos ^{-1}\left(\frac{a^2+b^2+c^2}{a b c}\right)\)

TS EAMCET-2018-05.05.2018

Vector Algebra

87705 \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) are non-coplanar vectors. If the position vector of the point of intersection of the line \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\mathbf{2} \overrightarrow{\mathrm{b}}+\mathbf{p}(\overrightarrow{\mathbf{a}}-\mathbf{2} \mathbf{c})\) and the plane \(\overrightarrow{\mathbf{r}}=3 \overrightarrow{\mathbf{a}}-\mathrm{q}(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathrm{b}})+K(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathbf{c}})\) is \(\overrightarrow{\mathbf{r}}=\mathbf{x} \overrightarrow{\mathbf{a}}+\mathbf{y} \vec{b}+z \overrightarrow{\mathbf{c}}\), then \(x\) y \(z\)

1 -8

2 8

3 12

4 -12

TS EAMCET-2021-05.08.2021

Vector Algebra

87706 If \(\vec{a}, \vec{b}, \vec{c}\) are the position vectors of the points A, B, C respectively, then match the items of list-I with those of list-II.
| List-I |List-II|
|
|A. \(\vec{a} =2 \hat{i}+3 \hat{j}+4 \hat{k}\),\ltbr>\(\vec{b} =3 \hat{i}+4 \hat{j}+2 \hat{k}\),\ltbr>\(\vec{c} =4 \hat{i}+2 \hat{j}+3 \hat{k}\)|I. A, B, C are collinear|
|B. \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}}=-3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\)|II. \(\triangle \mathrm{ABC}\) is an isosceles triangle|
|C. \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}}=-3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}\)|III. \(\triangle \mathrm{ABC}\) is a right angled triangle|
|D. \(\overrightarrow{\mathbf{a}} =\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}} =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}} =2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)|IV. \(\triangle \text { ABC}\) is a right-angled triangle
| | V. \(\Delta \text { ABC is }\),equilateral, triangle|
The correct match is

1 A-I, B-IV, C-III, D-II

2 A-I, B-II, C-III, D-IV

3 A-V, B-I, C-IV, D-II

4 A-V, B-I, C-III, D-II

Explanation:

(D) : Given position vectors are-
\(\overrightarrow{\mathrm{OA}}=\mathrm{a}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \Rightarrow \overrightarrow{\mathrm{OC}}=\mathrm{c}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{AB}}|=\sqrt{1^2+1^2+2^2}=\sqrt{6}\)
\(|\overrightarrow{\mathrm{BC}}|=\sqrt{1^2+2^2+1^2}=\sqrt{6}\)
And, \(|\overrightarrow{\mathrm{CA}}|=\sqrt{2^2+1^2+1^2}=\sqrt{6}\)
\(\therefore \triangle \mathrm{ABC}\) is an equilateral triangle. \(\mathrm{B}\), Given position vectors are
\(\overrightarrow{\mathrm{OA}}=\mathrm{a}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OC}}=\mathrm{c}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\)
\(\mid \overrightarrow{\mathrm{AB}}=\sqrt{4+4+16}=\sqrt{24}=2 \sqrt{6}\)
\(\mid \overrightarrow{\mathrm{BC}}=\sqrt{36+36+144}=216=6 \sqrt{6}\)
And,
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{16+16+64}=\sqrt{96}=4 \sqrt{6}\)
\(\because|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{CA}}|=|\overrightarrow{\mathrm{BC}}| \Rightarrow \mathrm{A}, \mathrm{B}, \mathrm{C}\) are collinear points.
C. given position vectors are
\(\overrightarrow{\mathrm{OA}}=a=2 \hat{i}-\hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{OB}}=b=\hat{i}-3 \hat{j}-5 \hat{k}\)
And,
\(\overrightarrow{\mathrm{OC}}=\mathrm{c}=-3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\)
\(\mid \overrightarrow{\mathrm{AB}}=\sqrt{1+4+36}=\sqrt{41}\)
\(\mid \overrightarrow{\mathrm{BC}}=\sqrt{16+1+1}=\sqrt{18}\)
And,
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{25+9+25}=\sqrt{59}\)
\(\because|\overrightarrow{\mathrm{CA}}|^2=|\overrightarrow{\mathrm{AB}}|^2+|\overrightarrow{\mathrm{BC}}|^2=\Delta \mathrm{ABC}\) is right angled triangle.
D. Given position sectors, \(\overrightarrow{O A}=a=\hat{i}+\hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{OC}}=\mathrm{c}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\therefore|\overrightarrow{\mathrm{AB}}|=\sqrt{0+1+4}=\sqrt{5}, \mid \overrightarrow{\mathrm{BC}}=\sqrt{1+9+4}=\sqrt{14}\)
And, \(|\overrightarrow{\mathrm{AC}}|=\sqrt{1+4+0}=\sqrt{5}\)
\(\because|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{BC}}| \text { and }|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{AC}}|>|\overrightarrow{\mathrm{BC}}|\)
So, \(\triangle A B C\) is isosceles triangle.

TS EAMCET-2020-11.09.2020

Vector Algebra

87707 If \(\vec{a}, \vec{b}, \vec{c}\) are three independent vectors and there exists a non zero scalar traid \((l, \mathrm{~m}, \mathrm{n})\) such that \(l(3 a+2 b+c)+\mathbf{m}(2 a+2 b+3 c)+\mathbf{n}(\mathbf{a}\) \(+2 b+5 c)=0\), then

1 \(l=\mathrm{m}=\mathrm{n}\)

2 \(l=\mathrm{n}\)

3 \(l=\mathrm{n}, \mathrm{m}+2 \mathrm{n}=0\)

4 \(\mathrm{m}+2 \mathrm{n}=0, l+\mathrm{n}=0\)

TS EAMCET-2020-14.09.2020

Vector Algebra

87703 If \(3 \hat{i}-5 \hat{j}+2 \hat{k}, 7 \hat{i}+2 \hat{j}-4 \hat{k}, \hat{i}-3 \hat{j}+4 \hat{k}\) and \(-7 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}\) are position vectors of the points \(A, B, C\) and \(D\) respectively, then the angle between \(A B\) and \(C D\) is

1 0

2 \(\frac{\pi}{4}\)

3 \(\frac{\pi}{2}\)

4 \(\pi\)

TS EAMCET-2022-18.07.2022

Vector Algebra

87704 If \(a, b, c\) are distinct real numbers and \(P, Q, R\) are three points whose position vectors are respectively \(\quad \mathbf{a} \hat{i}+b \hat{j}+c \hat{k}, b \hat{i}+c \hat{j}+a \hat{k} \quad\) and \(\hat{\mathbf{i}}+\mathbf{a} \hat{\mathbf{j}}+\mathbf{b} \hat{\mathbf{k}}\), then \(\angle \mathrm{QPR}=\)

1 \(\cos ^{-1}(a+b+c)\)

2 \(\frac{\pi}{2}\)

3 \(\frac{\pi}{3}\)

4 \(\cos ^{-1}\left(\frac{a^2+b^2+c^2}{a b c}\right)\)

TS EAMCET-2018-05.05.2018

Vector Algebra

87705 \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) are non-coplanar vectors. If the position vector of the point of intersection of the line \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\mathbf{2} \overrightarrow{\mathrm{b}}+\mathbf{p}(\overrightarrow{\mathbf{a}}-\mathbf{2} \mathbf{c})\) and the plane \(\overrightarrow{\mathbf{r}}=3 \overrightarrow{\mathbf{a}}-\mathrm{q}(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathrm{b}})+K(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathbf{c}})\) is \(\overrightarrow{\mathbf{r}}=\mathbf{x} \overrightarrow{\mathbf{a}}+\mathbf{y} \vec{b}+z \overrightarrow{\mathbf{c}}\), then \(x\) y \(z\)

1 -8

2 8

3 12

4 -12

TS EAMCET-2021-05.08.2021

Vector Algebra

87706 If \(\vec{a}, \vec{b}, \vec{c}\) are the position vectors of the points A, B, C respectively, then match the items of list-I with those of list-II.
| List-I |List-II|
|
|A. \(\vec{a} =2 \hat{i}+3 \hat{j}+4 \hat{k}\),\ltbr>\(\vec{b} =3 \hat{i}+4 \hat{j}+2 \hat{k}\),\ltbr>\(\vec{c} =4 \hat{i}+2 \hat{j}+3 \hat{k}\)|I. A, B, C are collinear|
|B. \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}}=-3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\)|II. \(\triangle \mathrm{ABC}\) is an isosceles triangle|
|C. \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}}=-3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}\)|III. \(\triangle \mathrm{ABC}\) is a right angled triangle|
|D. \(\overrightarrow{\mathbf{a}} =\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}} =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}} =2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)|IV. \(\triangle \text { ABC}\) is a right-angled triangle
| | V. \(\Delta \text { ABC is }\),equilateral, triangle|
The correct match is

1 A-I, B-IV, C-III, D-II

2 A-I, B-II, C-III, D-IV

3 A-V, B-I, C-IV, D-II

4 A-V, B-I, C-III, D-II

Explanation:

(D) : Given position vectors are-
\(\overrightarrow{\mathrm{OA}}=\mathrm{a}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \Rightarrow \overrightarrow{\mathrm{OC}}=\mathrm{c}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{AB}}|=\sqrt{1^2+1^2+2^2}=\sqrt{6}\)
\(|\overrightarrow{\mathrm{BC}}|=\sqrt{1^2+2^2+1^2}=\sqrt{6}\)
And, \(|\overrightarrow{\mathrm{CA}}|=\sqrt{2^2+1^2+1^2}=\sqrt{6}\)
\(\therefore \triangle \mathrm{ABC}\) is an equilateral triangle. \(\mathrm{B}\), Given position vectors are
\(\overrightarrow{\mathrm{OA}}=\mathrm{a}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OC}}=\mathrm{c}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\)
\(\mid \overrightarrow{\mathrm{AB}}=\sqrt{4+4+16}=\sqrt{24}=2 \sqrt{6}\)
\(\mid \overrightarrow{\mathrm{BC}}=\sqrt{36+36+144}=216=6 \sqrt{6}\)
And,
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{16+16+64}=\sqrt{96}=4 \sqrt{6}\)
\(\because|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{CA}}|=|\overrightarrow{\mathrm{BC}}| \Rightarrow \mathrm{A}, \mathrm{B}, \mathrm{C}\) are collinear points.
C. given position vectors are
\(\overrightarrow{\mathrm{OA}}=a=2 \hat{i}-\hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{OB}}=b=\hat{i}-3 \hat{j}-5 \hat{k}\)
And,
\(\overrightarrow{\mathrm{OC}}=\mathrm{c}=-3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\)
\(\mid \overrightarrow{\mathrm{AB}}=\sqrt{1+4+36}=\sqrt{41}\)
\(\mid \overrightarrow{\mathrm{BC}}=\sqrt{16+1+1}=\sqrt{18}\)
And,
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{25+9+25}=\sqrt{59}\)
\(\because|\overrightarrow{\mathrm{CA}}|^2=|\overrightarrow{\mathrm{AB}}|^2+|\overrightarrow{\mathrm{BC}}|^2=\Delta \mathrm{ABC}\) is right angled triangle.
D. Given position sectors, \(\overrightarrow{O A}=a=\hat{i}+\hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{OC}}=\mathrm{c}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\therefore|\overrightarrow{\mathrm{AB}}|=\sqrt{0+1+4}=\sqrt{5}, \mid \overrightarrow{\mathrm{BC}}=\sqrt{1+9+4}=\sqrt{14}\)
And, \(|\overrightarrow{\mathrm{AC}}|=\sqrt{1+4+0}=\sqrt{5}\)
\(\because|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{BC}}| \text { and }|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{AC}}|>|\overrightarrow{\mathrm{BC}}|\)
So, \(\triangle A B C\) is isosceles triangle.

TS EAMCET-2020-11.09.2020

Vector Algebra

87707 If \(\vec{a}, \vec{b}, \vec{c}\) are three independent vectors and there exists a non zero scalar traid \((l, \mathrm{~m}, \mathrm{n})\) such that \(l(3 a+2 b+c)+\mathbf{m}(2 a+2 b+3 c)+\mathbf{n}(\mathbf{a}\) \(+2 b+5 c)=0\), then

1 \(l=\mathrm{m}=\mathrm{n}\)

2 \(l=\mathrm{n}\)

3 \(l=\mathrm{n}, \mathrm{m}+2 \mathrm{n}=0\)

4 \(\mathrm{m}+2 \mathrm{n}=0, l+\mathrm{n}=0\)

TS EAMCET-2020-14.09.2020

Vector Algebra

87703 If \(3 \hat{i}-5 \hat{j}+2 \hat{k}, 7 \hat{i}+2 \hat{j}-4 \hat{k}, \hat{i}-3 \hat{j}+4 \hat{k}\) and \(-7 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}\) are position vectors of the points \(A, B, C\) and \(D\) respectively, then the angle between \(A B\) and \(C D\) is

1 0

2 \(\frac{\pi}{4}\)

3 \(\frac{\pi}{2}\)

4 \(\pi\)

TS EAMCET-2022-18.07.2022

Vector Algebra

87704 If \(a, b, c\) are distinct real numbers and \(P, Q, R\) are three points whose position vectors are respectively \(\quad \mathbf{a} \hat{i}+b \hat{j}+c \hat{k}, b \hat{i}+c \hat{j}+a \hat{k} \quad\) and \(\hat{\mathbf{i}}+\mathbf{a} \hat{\mathbf{j}}+\mathbf{b} \hat{\mathbf{k}}\), then \(\angle \mathrm{QPR}=\)

1 \(\cos ^{-1}(a+b+c)\)

2 \(\frac{\pi}{2}\)

3 \(\frac{\pi}{3}\)

4 \(\cos ^{-1}\left(\frac{a^2+b^2+c^2}{a b c}\right)\)

TS EAMCET-2018-05.05.2018

Vector Algebra

87705 \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) are non-coplanar vectors. If the position vector of the point of intersection of the line \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\mathbf{2} \overrightarrow{\mathrm{b}}+\mathbf{p}(\overrightarrow{\mathbf{a}}-\mathbf{2} \mathbf{c})\) and the plane \(\overrightarrow{\mathbf{r}}=3 \overrightarrow{\mathbf{a}}-\mathrm{q}(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathrm{b}})+K(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathbf{c}})\) is \(\overrightarrow{\mathbf{r}}=\mathbf{x} \overrightarrow{\mathbf{a}}+\mathbf{y} \vec{b}+z \overrightarrow{\mathbf{c}}\), then \(x\) y \(z\)

1 -8

2 8

3 12

4 -12

TS EAMCET-2021-05.08.2021

Vector Algebra

87706 If \(\vec{a}, \vec{b}, \vec{c}\) are the position vectors of the points A, B, C respectively, then match the items of list-I with those of list-II.
| List-I |List-II|
|
|A. \(\vec{a} =2 \hat{i}+3 \hat{j}+4 \hat{k}\),\ltbr>\(\vec{b} =3 \hat{i}+4 \hat{j}+2 \hat{k}\),\ltbr>\(\vec{c} =4 \hat{i}+2 \hat{j}+3 \hat{k}\)|I. A, B, C are collinear|
|B. \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}}=-3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\)|II. \(\triangle \mathrm{ABC}\) is an isosceles triangle|
|C. \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}}=-3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}\)|III. \(\triangle \mathrm{ABC}\) is a right angled triangle|
|D. \(\overrightarrow{\mathbf{a}} =\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}} =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}} =2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)|IV. \(\triangle \text { ABC}\) is a right-angled triangle
| | V. \(\Delta \text { ABC is }\),equilateral, triangle|
The correct match is

1 A-I, B-IV, C-III, D-II

2 A-I, B-II, C-III, D-IV

3 A-V, B-I, C-IV, D-II

4 A-V, B-I, C-III, D-II

Explanation:

(D) : Given position vectors are-
\(\overrightarrow{\mathrm{OA}}=\mathrm{a}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \Rightarrow \overrightarrow{\mathrm{OC}}=\mathrm{c}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{AB}}|=\sqrt{1^2+1^2+2^2}=\sqrt{6}\)
\(|\overrightarrow{\mathrm{BC}}|=\sqrt{1^2+2^2+1^2}=\sqrt{6}\)
And, \(|\overrightarrow{\mathrm{CA}}|=\sqrt{2^2+1^2+1^2}=\sqrt{6}\)
\(\therefore \triangle \mathrm{ABC}\) is an equilateral triangle. \(\mathrm{B}\), Given position vectors are
\(\overrightarrow{\mathrm{OA}}=\mathrm{a}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OC}}=\mathrm{c}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\)
\(\mid \overrightarrow{\mathrm{AB}}=\sqrt{4+4+16}=\sqrt{24}=2 \sqrt{6}\)
\(\mid \overrightarrow{\mathrm{BC}}=\sqrt{36+36+144}=216=6 \sqrt{6}\)
And,
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{16+16+64}=\sqrt{96}=4 \sqrt{6}\)
\(\because|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{CA}}|=|\overrightarrow{\mathrm{BC}}| \Rightarrow \mathrm{A}, \mathrm{B}, \mathrm{C}\) are collinear points.
C. given position vectors are
\(\overrightarrow{\mathrm{OA}}=a=2 \hat{i}-\hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{OB}}=b=\hat{i}-3 \hat{j}-5 \hat{k}\)
And,
\(\overrightarrow{\mathrm{OC}}=\mathrm{c}=-3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\)
\(\mid \overrightarrow{\mathrm{AB}}=\sqrt{1+4+36}=\sqrt{41}\)
\(\mid \overrightarrow{\mathrm{BC}}=\sqrt{16+1+1}=\sqrt{18}\)
And,
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{25+9+25}=\sqrt{59}\)
\(\because|\overrightarrow{\mathrm{CA}}|^2=|\overrightarrow{\mathrm{AB}}|^2+|\overrightarrow{\mathrm{BC}}|^2=\Delta \mathrm{ABC}\) is right angled triangle.
D. Given position sectors, \(\overrightarrow{O A}=a=\hat{i}+\hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{OC}}=\mathrm{c}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\therefore|\overrightarrow{\mathrm{AB}}|=\sqrt{0+1+4}=\sqrt{5}, \mid \overrightarrow{\mathrm{BC}}=\sqrt{1+9+4}=\sqrt{14}\)
And, \(|\overrightarrow{\mathrm{AC}}|=\sqrt{1+4+0}=\sqrt{5}\)
\(\because|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{BC}}| \text { and }|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{AC}}|>|\overrightarrow{\mathrm{BC}}|\)
So, \(\triangle A B C\) is isosceles triangle.

TS EAMCET-2020-11.09.2020

Vector Algebra

87707 If \(\vec{a}, \vec{b}, \vec{c}\) are three independent vectors and there exists a non zero scalar traid \((l, \mathrm{~m}, \mathrm{n})\) such that \(l(3 a+2 b+c)+\mathbf{m}(2 a+2 b+3 c)+\mathbf{n}(\mathbf{a}\) \(+2 b+5 c)=0\), then

1 \(l=\mathrm{m}=\mathrm{n}\)

2 \(l=\mathrm{n}\)

3 \(l=\mathrm{n}, \mathrm{m}+2 \mathrm{n}=0\)

4 \(\mathrm{m}+2 \mathrm{n}=0, l+\mathrm{n}=0\)

TS EAMCET-2020-14.09.2020

Vector Algebra

87703 If \(3 \hat{i}-5 \hat{j}+2 \hat{k}, 7 \hat{i}+2 \hat{j}-4 \hat{k}, \hat{i}-3 \hat{j}+4 \hat{k}\) and \(-7 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}\) are position vectors of the points \(A, B, C\) and \(D\) respectively, then the angle between \(A B\) and \(C D\) is

1 0

2 \(\frac{\pi}{4}\)

3 \(\frac{\pi}{2}\)

4 \(\pi\)

TS EAMCET-2022-18.07.2022

Vector Algebra

87704 If \(a, b, c\) are distinct real numbers and \(P, Q, R\) are three points whose position vectors are respectively \(\quad \mathbf{a} \hat{i}+b \hat{j}+c \hat{k}, b \hat{i}+c \hat{j}+a \hat{k} \quad\) and \(\hat{\mathbf{i}}+\mathbf{a} \hat{\mathbf{j}}+\mathbf{b} \hat{\mathbf{k}}\), then \(\angle \mathrm{QPR}=\)

1 \(\cos ^{-1}(a+b+c)\)

2 \(\frac{\pi}{2}\)

3 \(\frac{\pi}{3}\)

4 \(\cos ^{-1}\left(\frac{a^2+b^2+c^2}{a b c}\right)\)

TS EAMCET-2018-05.05.2018

Vector Algebra

87705 \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\) are non-coplanar vectors. If the position vector of the point of intersection of the line \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\mathbf{2} \overrightarrow{\mathrm{b}}+\mathbf{p}(\overrightarrow{\mathbf{a}}-\mathbf{2} \mathbf{c})\) and the plane \(\overrightarrow{\mathbf{r}}=3 \overrightarrow{\mathbf{a}}-\mathrm{q}(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathrm{b}})+K(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathbf{c}})\) is \(\overrightarrow{\mathbf{r}}=\mathbf{x} \overrightarrow{\mathbf{a}}+\mathbf{y} \vec{b}+z \overrightarrow{\mathbf{c}}\), then \(x\) y \(z\)

1 -8

2 8

3 12

4 -12

TS EAMCET-2021-05.08.2021

Vector Algebra

87706 If \(\vec{a}, \vec{b}, \vec{c}\) are the position vectors of the points A, B, C respectively, then match the items of list-I with those of list-II.
| List-I |List-II|
|
|A. \(\vec{a} =2 \hat{i}+3 \hat{j}+4 \hat{k}\),\ltbr>\(\vec{b} =3 \hat{i}+4 \hat{j}+2 \hat{k}\),\ltbr>\(\vec{c} =4 \hat{i}+2 \hat{j}+3 \hat{k}\)|I. A, B, C are collinear|
|B. \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}}=-3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\)|II. \(\triangle \mathrm{ABC}\) is an isosceles triangle|
|C. \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\mathbf{5} \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}}=-3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}\)|III. \(\triangle \mathrm{ABC}\) is a right angled triangle|
|D. \(\overrightarrow{\mathbf{a}} =\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{b}} =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\),\ltbr>\(\overrightarrow{\mathbf{c}} =2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)|IV. \(\triangle \text { ABC}\) is a right-angled triangle
| | V. \(\Delta \text { ABC is }\),equilateral, triangle|
The correct match is

1 A-I, B-IV, C-III, D-II

2 A-I, B-II, C-III, D-IV

3 A-V, B-I, C-IV, D-II

4 A-V, B-I, C-III, D-II

Explanation:

(D) : Given position vectors are-
\(\overrightarrow{\mathrm{OA}}=\mathrm{a}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \Rightarrow \overrightarrow{\mathrm{OC}}=\mathrm{c}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(|\overrightarrow{\mathrm{AB}}|=\sqrt{1^2+1^2+2^2}=\sqrt{6}\)
\(|\overrightarrow{\mathrm{BC}}|=\sqrt{1^2+2^2+1^2}=\sqrt{6}\)
And, \(|\overrightarrow{\mathrm{CA}}|=\sqrt{2^2+1^2+1^2}=\sqrt{6}\)
\(\therefore \triangle \mathrm{ABC}\) is an equilateral triangle. \(\mathrm{B}\), Given position vectors are
\(\overrightarrow{\mathrm{OA}}=\mathrm{a}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{OC}}=\mathrm{c}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\)
\(\mid \overrightarrow{\mathrm{AB}}=\sqrt{4+4+16}=\sqrt{24}=2 \sqrt{6}\)
\(\mid \overrightarrow{\mathrm{BC}}=\sqrt{36+36+144}=216=6 \sqrt{6}\)
And,
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{16+16+64}=\sqrt{96}=4 \sqrt{6}\)
\(\because|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{CA}}|=|\overrightarrow{\mathrm{BC}}| \Rightarrow \mathrm{A}, \mathrm{B}, \mathrm{C}\) are collinear points.
C. given position vectors are
\(\overrightarrow{\mathrm{OA}}=a=2 \hat{i}-\hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{OB}}=b=\hat{i}-3 \hat{j}-5 \hat{k}\)
And,
\(\overrightarrow{\mathrm{OC}}=\mathrm{c}=-3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\)
\(\mid \overrightarrow{\mathrm{AB}}=\sqrt{1+4+36}=\sqrt{41}\)
\(\mid \overrightarrow{\mathrm{BC}}=\sqrt{16+1+1}=\sqrt{18}\)
And,
\(|\overrightarrow{\mathrm{CA}}|=\sqrt{25+9+25}=\sqrt{59}\)
\(\because|\overrightarrow{\mathrm{CA}}|^2=|\overrightarrow{\mathrm{AB}}|^2+|\overrightarrow{\mathrm{BC}}|^2=\Delta \mathrm{ABC}\) is right angled triangle.
D. Given position sectors, \(\overrightarrow{O A}=a=\hat{i}+\hat{j}+\hat{k}\)
\(\overrightarrow{\mathrm{OB}}=\mathrm{b}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{OC}}=\mathrm{c}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\therefore|\overrightarrow{\mathrm{AB}}|=\sqrt{0+1+4}=\sqrt{5}, \mid \overrightarrow{\mathrm{BC}}=\sqrt{1+9+4}=\sqrt{14}\)
And, \(|\overrightarrow{\mathrm{AC}}|=\sqrt{1+4+0}=\sqrt{5}\)
\(\because|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{BC}}| \text { and }|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{AC}}|>|\overrightarrow{\mathrm{BC}}|\)
So, \(\triangle A B C\) is isosceles triangle.

TS EAMCET-2020-11.09.2020

Vector Algebra

87707 If \(\vec{a}, \vec{b}, \vec{c}\) are three independent vectors and there exists a non zero scalar traid \((l, \mathrm{~m}, \mathrm{n})\) such that \(l(3 a+2 b+c)+\mathbf{m}(2 a+2 b+3 c)+\mathbf{n}(\mathbf{a}\) \(+2 b+5 c)=0\), then

1 \(l=\mathrm{m}=\mathrm{n}\)

2 \(l=\mathrm{n}\)

3 \(l=\mathrm{n}, \mathrm{m}+2 \mathrm{n}=0\)

4 \(\mathrm{m}+2 \mathrm{n}=0, l+\mathrm{n}=0\)

TS EAMCET-2020-14.09.2020

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