QBManager

Vector Algebra

87712 Let $A (3 \hat{i} + \hat{j} - \hat{k})$ and $B (13 \hat{i} - 4 \hat{j} + 9 \hat{k})$ be two points on a line $L$ . $C$ and $D$ be the points on $L$ on either side of $A$ at distance of 9 and 6 units respectively and $C$ lies between $A$ and $B$ . Then position vectors of $C$ and $D$ are respectively

1

- 3 \hat{i} + 4 \hat{j} - 7 \hat{k}, 7 \hat{i} - \hat{j} + 3 \hat{k}

2

9 \hat{i} - 2 \hat{j} + 5 \hat{k}, 7 \hat{i} - \hat{j} + 3 \hat{k}

3

- 3 \hat{i} + 4 \hat{j} - 7 \hat{k}, - \hat{i} + 3 \hat{j} - 5 \hat{k}

4

9 \hat{i} - 2 \hat{j} + 5 \hat{k}, - \hat{i} + 3 \hat{j} - 5 \hat{k}

Explanation:

(D) : Given, $A (3 \hat{i} - \hat{j} - \hat{k})$ and $B (13 \hat{i} - 4 \hat{j} + 9 \hat{k})$
$AB = \sqrt{10^{2} + 5^{2} + 10^{2}} = 15$
original image
$AC = 9, DA = 6$
$∵ C = \frac{9 (13 \hat{i} - 4 \hat{j} + 9 \hat{k}) + 6 (3 \hat{i} + \hat{j} - \hat{k})}{9 + 6}$
$C = \frac{135 \hat{i} - 30 \hat{j} + 75 \hat{k}}{15} = 9 \hat{i} - 2 \hat{j} + 5 \hat{k}$
Let position vector of $D$ is $x \hat{i} + y \hat{j} + z \hat{k}$
$∴ 3 \hat{i} + \hat{j} - \hat{k} = \frac{6 \times (9 \hat{i} - 2 \hat{j} + 5 \hat{k}) + 9 (x \hat{i} + y \hat{j} + z \hat{k})}{6 + 9}$
$\Rightarrow 3 \hat{i} + \hat{j} - \hat{k} = \frac{2 (9 \hat{i} - 2 \hat{j} + 5 \hat{k}) + 3 (x \hat{i} + y \hat{j} + z \hat{k})}{5}$
$\Rightarrow 15 \hat{i} + 5 \hat{j} - 5 \hat{k} = (18 + 3 x) \hat{i} + (- 4 + 3 y) \hat{j} + (10 + 3 z) \hat{k}$
$∴ 15 = 18 + 3 x \Rightarrow x = - 1$
$5 = - 4 + 3 y \Rightarrow y = 3$
$And, - 5 = 10 + 3 z \Rightarrow z = - 5$
$∴ Position vector of D is - \hat{i} + 3 \hat{j} - 5 \hat{k}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87713 If the orthocentre of the triangle whose vertices are $2 \hat{i} + 3 \hat{j} + 5 \hat{k}, 5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $3 \hat{i} + 5 \hat{j} + 2 \hat{k}$ is $x + y + z \hat{k}$ , then

1

x = 2 y = z

2

x = y = 2 z

3

x = y = - z

4

x = y = z

Explanation:

(D) : The vertices of triangle are $2 \hat{i} + 3 \hat{j} + 5 \hat{k}, 5 \hat{i} + 2 \hat{j} + 3 \hat{k}, 3 \hat{i} + 5 \hat{j} + 2 \hat{k}$
original image
$\overset{―}{AB} = 3 \hat{i} - \hat{j} - 2 \hat{k}$
$\overset{―}{BC} = - 2 \hat{i} + 3 \hat{j} - \hat{k}$
$\vec{AC} = \hat{i} + 2 \hat{j} - 3 \hat{k}$
$| \vec{AB} | = \sqrt{9 + 1 + 4} = \sqrt{14}$
$| \vec{BC} | = \sqrt{4 + 9 + 1} = \sqrt{14}$
$| \vec{AC} | = \sqrt{1 + 4 + 9} = \sqrt{14}$
$△ ABC$ is an equilateral triangle
$∴$ Orthocentre of triangle is coincide with centroid
$∴$ Orthocentre of $△ ABC$ is
$(\frac{2 \hat{i} + 3 \hat{j} + 5 \hat{k} + 5 \hat{i} + 2 \hat{j} + 3 \hat{k} + 3 \hat{i} + 5 \hat{j} + 2 \hat{k}}{3})$
$= \frac{15 \hat{i} + 15 \hat{j} + 15 \hat{k}}{3} = 5 \hat{i} + 5 \hat{j} + 5 \hat{k}$
So, $x = y = z = 5$

TS EAMCET-2020-10.09.2020

Vector Algebra

87714 If $P, Q, R$ are the mid-point of the sides $A B, B C$ and $C A$ of $△ A B C$ respectively, then $P C - B Q =$

1

CP

2

PQ

3

BR

4

AR

Explanation:

, d) : Let position vectors of vertices A,B,C are $a, b, c$ respectively
Since, $P, Q, R$ are mid points of $A B, B C, C A$
original image
$∴$ Position vector of $P, Q, R$ are
$(\frac{a + b}{2}) (\frac{b + c}{2}) (\frac{c + a}{2})$
Respectively
Now,
$PC - BQ = [c - (\frac{a + b}{2})] - [(\frac{b + c}{2}) - b]$
$= \frac{2 c - a - b}{2} - \frac{b + c - 2 b}{2} = \frac{2 c - a - b - b - c + 2 b}{2} = \frac{c - a}{2}$
Now,
$CP = \frac{a + b}{2} - c = \frac{a + b + 2 c}{2}$
$PQ = \frac{b + c}{2} - \frac{a + b}{2} = \frac{c - a}{2}$
$BR = \frac{c + a}{2} - b = \frac{c - a - 2 b}{2}$
$AR = \frac{c + a}{2} - a = \frac{c - a}{2}$
$∴ PC - BQ = PQ = AR$

TS EAMCET-2018-07.05.2018

Vector Algebra

87715 The position vector of $a$ point $P$ is $2 \hat{i} + \hat{j} + 3 \hat{k}$ and $a = - \hat{i} - 2 \hat{k}, b = \hat{i} + \hat{j} + 2 \hat{k}$ are two vectors which determine a plane $π$ . The equation of a line through $P$ normal to $b$ and lying on the plane $π$ is

1

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- \hat{i} + 5 \hat{j} - 2 \hat{k})

2

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (\hat{i} + \hat{j} + \hat{k})

3

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- 2 \hat{i} - \hat{j} + 3 \hat{k})

4

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- 3 \hat{i} + 4 \hat{j} - 5 \hat{k})

Explanation:

(A) : Given,
Position vector of a point $P$ is $(2 \hat{i} + \hat{j} + 3 \hat{k})$
Equation of plane,
$(r - 2 \hat{i} + \hat{j} + 3 \hat{k}) \cdot (a \times b) = 0$
$∴$ Normal of plane $a \times b$
Equation of line through $P (2 \hat{i} + \hat{j} + 3 \hat{k})$ and normal to $b$ and lying on the plane is
$r = (2 \hat{i} + \hat{j} + 3 \hat{k}) + λ {a \times (b \times c)}$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ {(b \cdot a) b - (b \cdot b) a}$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ [(- 5) (\hat{i} + \hat{j} + 2 \hat{k}) - 6 (- \hat{i} - 2 \hat{k})]$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- \hat{i} + 5 \hat{j} - 2 \hat{k})$

TS EAMCET-2020-10.09.2020

Vector Algebra

87712 Let $A (3 \hat{i} + \hat{j} - \hat{k})$ and $B (13 \hat{i} - 4 \hat{j} + 9 \hat{k})$ be two points on a line $L$ . $C$ and $D$ be the points on $L$ on either side of $A$ at distance of 9 and 6 units respectively and $C$ lies between $A$ and $B$ . Then position vectors of $C$ and $D$ are respectively

1

- 3 \hat{i} + 4 \hat{j} - 7 \hat{k}, 7 \hat{i} - \hat{j} + 3 \hat{k}

2

9 \hat{i} - 2 \hat{j} + 5 \hat{k}, 7 \hat{i} - \hat{j} + 3 \hat{k}

3

- 3 \hat{i} + 4 \hat{j} - 7 \hat{k}, - \hat{i} + 3 \hat{j} - 5 \hat{k}

4

9 \hat{i} - 2 \hat{j} + 5 \hat{k}, - \hat{i} + 3 \hat{j} - 5 \hat{k}

Explanation:

(D) : Given, $A (3 \hat{i} - \hat{j} - \hat{k})$ and $B (13 \hat{i} - 4 \hat{j} + 9 \hat{k})$
$AB = \sqrt{10^{2} + 5^{2} + 10^{2}} = 15$
original image
$AC = 9, DA = 6$
$∵ C = \frac{9 (13 \hat{i} - 4 \hat{j} + 9 \hat{k}) + 6 (3 \hat{i} + \hat{j} - \hat{k})}{9 + 6}$
$C = \frac{135 \hat{i} - 30 \hat{j} + 75 \hat{k}}{15} = 9 \hat{i} - 2 \hat{j} + 5 \hat{k}$
Let position vector of $D$ is $x \hat{i} + y \hat{j} + z \hat{k}$
$∴ 3 \hat{i} + \hat{j} - \hat{k} = \frac{6 \times (9 \hat{i} - 2 \hat{j} + 5 \hat{k}) + 9 (x \hat{i} + y \hat{j} + z \hat{k})}{6 + 9}$
$\Rightarrow 3 \hat{i} + \hat{j} - \hat{k} = \frac{2 (9 \hat{i} - 2 \hat{j} + 5 \hat{k}) + 3 (x \hat{i} + y \hat{j} + z \hat{k})}{5}$
$\Rightarrow 15 \hat{i} + 5 \hat{j} - 5 \hat{k} = (18 + 3 x) \hat{i} + (- 4 + 3 y) \hat{j} + (10 + 3 z) \hat{k}$
$∴ 15 = 18 + 3 x \Rightarrow x = - 1$
$5 = - 4 + 3 y \Rightarrow y = 3$
$And, - 5 = 10 + 3 z \Rightarrow z = - 5$
$∴ Position vector of D is - \hat{i} + 3 \hat{j} - 5 \hat{k}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87713 If the orthocentre of the triangle whose vertices are $2 \hat{i} + 3 \hat{j} + 5 \hat{k}, 5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $3 \hat{i} + 5 \hat{j} + 2 \hat{k}$ is $x + y + z \hat{k}$ , then

1

x = 2 y = z

2

x = y = 2 z

3

x = y = - z

4

x = y = z

Explanation:

(D) : The vertices of triangle are $2 \hat{i} + 3 \hat{j} + 5 \hat{k}, 5 \hat{i} + 2 \hat{j} + 3 \hat{k}, 3 \hat{i} + 5 \hat{j} + 2 \hat{k}$
original image
$\overset{―}{AB} = 3 \hat{i} - \hat{j} - 2 \hat{k}$
$\overset{―}{BC} = - 2 \hat{i} + 3 \hat{j} - \hat{k}$
$\vec{AC} = \hat{i} + 2 \hat{j} - 3 \hat{k}$
$| \vec{AB} | = \sqrt{9 + 1 + 4} = \sqrt{14}$
$| \vec{BC} | = \sqrt{4 + 9 + 1} = \sqrt{14}$
$| \vec{AC} | = \sqrt{1 + 4 + 9} = \sqrt{14}$
$△ ABC$ is an equilateral triangle
$∴$ Orthocentre of triangle is coincide with centroid
$∴$ Orthocentre of $△ ABC$ is
$(\frac{2 \hat{i} + 3 \hat{j} + 5 \hat{k} + 5 \hat{i} + 2 \hat{j} + 3 \hat{k} + 3 \hat{i} + 5 \hat{j} + 2 \hat{k}}{3})$
$= \frac{15 \hat{i} + 15 \hat{j} + 15 \hat{k}}{3} = 5 \hat{i} + 5 \hat{j} + 5 \hat{k}$
So, $x = y = z = 5$

TS EAMCET-2020-10.09.2020

Vector Algebra

87714 If $P, Q, R$ are the mid-point of the sides $A B, B C$ and $C A$ of $△ A B C$ respectively, then $P C - B Q =$

1

CP

2

PQ

3

BR

4

AR

Explanation:

, d) : Let position vectors of vertices A,B,C are $a, b, c$ respectively
Since, $P, Q, R$ are mid points of $A B, B C, C A$
original image
$∴$ Position vector of $P, Q, R$ are
$(\frac{a + b}{2}) (\frac{b + c}{2}) (\frac{c + a}{2})$
Respectively
Now,
$PC - BQ = [c - (\frac{a + b}{2})] - [(\frac{b + c}{2}) - b]$
$= \frac{2 c - a - b}{2} - \frac{b + c - 2 b}{2} = \frac{2 c - a - b - b - c + 2 b}{2} = \frac{c - a}{2}$
Now,
$CP = \frac{a + b}{2} - c = \frac{a + b + 2 c}{2}$
$PQ = \frac{b + c}{2} - \frac{a + b}{2} = \frac{c - a}{2}$
$BR = \frac{c + a}{2} - b = \frac{c - a - 2 b}{2}$
$AR = \frac{c + a}{2} - a = \frac{c - a}{2}$
$∴ PC - BQ = PQ = AR$

TS EAMCET-2018-07.05.2018

Vector Algebra

87715 The position vector of $a$ point $P$ is $2 \hat{i} + \hat{j} + 3 \hat{k}$ and $a = - \hat{i} - 2 \hat{k}, b = \hat{i} + \hat{j} + 2 \hat{k}$ are two vectors which determine a plane $π$ . The equation of a line through $P$ normal to $b$ and lying on the plane $π$ is

1

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- \hat{i} + 5 \hat{j} - 2 \hat{k})

2

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (\hat{i} + \hat{j} + \hat{k})

3

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- 2 \hat{i} - \hat{j} + 3 \hat{k})

4

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- 3 \hat{i} + 4 \hat{j} - 5 \hat{k})

Explanation:

(A) : Given,
Position vector of a point $P$ is $(2 \hat{i} + \hat{j} + 3 \hat{k})$
Equation of plane,
$(r - 2 \hat{i} + \hat{j} + 3 \hat{k}) \cdot (a \times b) = 0$
$∴$ Normal of plane $a \times b$
Equation of line through $P (2 \hat{i} + \hat{j} + 3 \hat{k})$ and normal to $b$ and lying on the plane is
$r = (2 \hat{i} + \hat{j} + 3 \hat{k}) + λ {a \times (b \times c)}$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ {(b \cdot a) b - (b \cdot b) a}$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ [(- 5) (\hat{i} + \hat{j} + 2 \hat{k}) - 6 (- \hat{i} - 2 \hat{k})]$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- \hat{i} + 5 \hat{j} - 2 \hat{k})$

TS EAMCET-2020-10.09.2020

Vector Algebra

87712 Let $A (3 \hat{i} + \hat{j} - \hat{k})$ and $B (13 \hat{i} - 4 \hat{j} + 9 \hat{k})$ be two points on a line $L$ . $C$ and $D$ be the points on $L$ on either side of $A$ at distance of 9 and 6 units respectively and $C$ lies between $A$ and $B$ . Then position vectors of $C$ and $D$ are respectively

1

- 3 \hat{i} + 4 \hat{j} - 7 \hat{k}, 7 \hat{i} - \hat{j} + 3 \hat{k}

2

9 \hat{i} - 2 \hat{j} + 5 \hat{k}, 7 \hat{i} - \hat{j} + 3 \hat{k}

3

- 3 \hat{i} + 4 \hat{j} - 7 \hat{k}, - \hat{i} + 3 \hat{j} - 5 \hat{k}

4

9 \hat{i} - 2 \hat{j} + 5 \hat{k}, - \hat{i} + 3 \hat{j} - 5 \hat{k}

Explanation:

(D) : Given, $A (3 \hat{i} - \hat{j} - \hat{k})$ and $B (13 \hat{i} - 4 \hat{j} + 9 \hat{k})$
$AB = \sqrt{10^{2} + 5^{2} + 10^{2}} = 15$
original image
$AC = 9, DA = 6$
$∵ C = \frac{9 (13 \hat{i} - 4 \hat{j} + 9 \hat{k}) + 6 (3 \hat{i} + \hat{j} - \hat{k})}{9 + 6}$
$C = \frac{135 \hat{i} - 30 \hat{j} + 75 \hat{k}}{15} = 9 \hat{i} - 2 \hat{j} + 5 \hat{k}$
Let position vector of $D$ is $x \hat{i} + y \hat{j} + z \hat{k}$
$∴ 3 \hat{i} + \hat{j} - \hat{k} = \frac{6 \times (9 \hat{i} - 2 \hat{j} + 5 \hat{k}) + 9 (x \hat{i} + y \hat{j} + z \hat{k})}{6 + 9}$
$\Rightarrow 3 \hat{i} + \hat{j} - \hat{k} = \frac{2 (9 \hat{i} - 2 \hat{j} + 5 \hat{k}) + 3 (x \hat{i} + y \hat{j} + z \hat{k})}{5}$
$\Rightarrow 15 \hat{i} + 5 \hat{j} - 5 \hat{k} = (18 + 3 x) \hat{i} + (- 4 + 3 y) \hat{j} + (10 + 3 z) \hat{k}$
$∴ 15 = 18 + 3 x \Rightarrow x = - 1$
$5 = - 4 + 3 y \Rightarrow y = 3$
$And, - 5 = 10 + 3 z \Rightarrow z = - 5$
$∴ Position vector of D is - \hat{i} + 3 \hat{j} - 5 \hat{k}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87713 If the orthocentre of the triangle whose vertices are $2 \hat{i} + 3 \hat{j} + 5 \hat{k}, 5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $3 \hat{i} + 5 \hat{j} + 2 \hat{k}$ is $x + y + z \hat{k}$ , then

1

x = 2 y = z

2

x = y = 2 z

3

x = y = - z

4

x = y = z

Explanation:

(D) : The vertices of triangle are $2 \hat{i} + 3 \hat{j} + 5 \hat{k}, 5 \hat{i} + 2 \hat{j} + 3 \hat{k}, 3 \hat{i} + 5 \hat{j} + 2 \hat{k}$
original image
$\overset{―}{AB} = 3 \hat{i} - \hat{j} - 2 \hat{k}$
$\overset{―}{BC} = - 2 \hat{i} + 3 \hat{j} - \hat{k}$
$\vec{AC} = \hat{i} + 2 \hat{j} - 3 \hat{k}$
$| \vec{AB} | = \sqrt{9 + 1 + 4} = \sqrt{14}$
$| \vec{BC} | = \sqrt{4 + 9 + 1} = \sqrt{14}$
$| \vec{AC} | = \sqrt{1 + 4 + 9} = \sqrt{14}$
$△ ABC$ is an equilateral triangle
$∴$ Orthocentre of triangle is coincide with centroid
$∴$ Orthocentre of $△ ABC$ is
$(\frac{2 \hat{i} + 3 \hat{j} + 5 \hat{k} + 5 \hat{i} + 2 \hat{j} + 3 \hat{k} + 3 \hat{i} + 5 \hat{j} + 2 \hat{k}}{3})$
$= \frac{15 \hat{i} + 15 \hat{j} + 15 \hat{k}}{3} = 5 \hat{i} + 5 \hat{j} + 5 \hat{k}$
So, $x = y = z = 5$

TS EAMCET-2020-10.09.2020

Vector Algebra

87714 If $P, Q, R$ are the mid-point of the sides $A B, B C$ and $C A$ of $△ A B C$ respectively, then $P C - B Q =$

1

CP

2

PQ

3

BR

4

AR

Explanation:

, d) : Let position vectors of vertices A,B,C are $a, b, c$ respectively
Since, $P, Q, R$ are mid points of $A B, B C, C A$
original image
$∴$ Position vector of $P, Q, R$ are
$(\frac{a + b}{2}) (\frac{b + c}{2}) (\frac{c + a}{2})$
Respectively
Now,
$PC - BQ = [c - (\frac{a + b}{2})] - [(\frac{b + c}{2}) - b]$
$= \frac{2 c - a - b}{2} - \frac{b + c - 2 b}{2} = \frac{2 c - a - b - b - c + 2 b}{2} = \frac{c - a}{2}$
Now,
$CP = \frac{a + b}{2} - c = \frac{a + b + 2 c}{2}$
$PQ = \frac{b + c}{2} - \frac{a + b}{2} = \frac{c - a}{2}$
$BR = \frac{c + a}{2} - b = \frac{c - a - 2 b}{2}$
$AR = \frac{c + a}{2} - a = \frac{c - a}{2}$
$∴ PC - BQ = PQ = AR$

TS EAMCET-2018-07.05.2018

Vector Algebra

87715 The position vector of $a$ point $P$ is $2 \hat{i} + \hat{j} + 3 \hat{k}$ and $a = - \hat{i} - 2 \hat{k}, b = \hat{i} + \hat{j} + 2 \hat{k}$ are two vectors which determine a plane $π$ . The equation of a line through $P$ normal to $b$ and lying on the plane $π$ is

1

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- \hat{i} + 5 \hat{j} - 2 \hat{k})

2

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (\hat{i} + \hat{j} + \hat{k})

3

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- 2 \hat{i} - \hat{j} + 3 \hat{k})

4

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- 3 \hat{i} + 4 \hat{j} - 5 \hat{k})

Explanation:

(A) : Given,
Position vector of a point $P$ is $(2 \hat{i} + \hat{j} + 3 \hat{k})$
Equation of plane,
$(r - 2 \hat{i} + \hat{j} + 3 \hat{k}) \cdot (a \times b) = 0$
$∴$ Normal of plane $a \times b$
Equation of line through $P (2 \hat{i} + \hat{j} + 3 \hat{k})$ and normal to $b$ and lying on the plane is
$r = (2 \hat{i} + \hat{j} + 3 \hat{k}) + λ {a \times (b \times c)}$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ {(b \cdot a) b - (b \cdot b) a}$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ [(- 5) (\hat{i} + \hat{j} + 2 \hat{k}) - 6 (- \hat{i} - 2 \hat{k})]$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- \hat{i} + 5 \hat{j} - 2 \hat{k})$

TS EAMCET-2020-10.09.2020

Vector Algebra

87712 Let $A (3 \hat{i} + \hat{j} - \hat{k})$ and $B (13 \hat{i} - 4 \hat{j} + 9 \hat{k})$ be two points on a line $L$ . $C$ and $D$ be the points on $L$ on either side of $A$ at distance of 9 and 6 units respectively and $C$ lies between $A$ and $B$ . Then position vectors of $C$ and $D$ are respectively

1

- 3 \hat{i} + 4 \hat{j} - 7 \hat{k}, 7 \hat{i} - \hat{j} + 3 \hat{k}

2

9 \hat{i} - 2 \hat{j} + 5 \hat{k}, 7 \hat{i} - \hat{j} + 3 \hat{k}

3

- 3 \hat{i} + 4 \hat{j} - 7 \hat{k}, - \hat{i} + 3 \hat{j} - 5 \hat{k}

4

9 \hat{i} - 2 \hat{j} + 5 \hat{k}, - \hat{i} + 3 \hat{j} - 5 \hat{k}

Explanation:

(D) : Given, $A (3 \hat{i} - \hat{j} - \hat{k})$ and $B (13 \hat{i} - 4 \hat{j} + 9 \hat{k})$
$AB = \sqrt{10^{2} + 5^{2} + 10^{2}} = 15$
original image
$AC = 9, DA = 6$
$∵ C = \frac{9 (13 \hat{i} - 4 \hat{j} + 9 \hat{k}) + 6 (3 \hat{i} + \hat{j} - \hat{k})}{9 + 6}$
$C = \frac{135 \hat{i} - 30 \hat{j} + 75 \hat{k}}{15} = 9 \hat{i} - 2 \hat{j} + 5 \hat{k}$
Let position vector of $D$ is $x \hat{i} + y \hat{j} + z \hat{k}$
$∴ 3 \hat{i} + \hat{j} - \hat{k} = \frac{6 \times (9 \hat{i} - 2 \hat{j} + 5 \hat{k}) + 9 (x \hat{i} + y \hat{j} + z \hat{k})}{6 + 9}$
$\Rightarrow 3 \hat{i} + \hat{j} - \hat{k} = \frac{2 (9 \hat{i} - 2 \hat{j} + 5 \hat{k}) + 3 (x \hat{i} + y \hat{j} + z \hat{k})}{5}$
$\Rightarrow 15 \hat{i} + 5 \hat{j} - 5 \hat{k} = (18 + 3 x) \hat{i} + (- 4 + 3 y) \hat{j} + (10 + 3 z) \hat{k}$
$∴ 15 = 18 + 3 x \Rightarrow x = - 1$
$5 = - 4 + 3 y \Rightarrow y = 3$
$And, - 5 = 10 + 3 z \Rightarrow z = - 5$
$∴ Position vector of D is - \hat{i} + 3 \hat{j} - 5 \hat{k}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87713 If the orthocentre of the triangle whose vertices are $2 \hat{i} + 3 \hat{j} + 5 \hat{k}, 5 \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $3 \hat{i} + 5 \hat{j} + 2 \hat{k}$ is $x + y + z \hat{k}$ , then

1

x = 2 y = z

2

x = y = 2 z

3

x = y = - z

4

x = y = z

Explanation:

(D) : The vertices of triangle are $2 \hat{i} + 3 \hat{j} + 5 \hat{k}, 5 \hat{i} + 2 \hat{j} + 3 \hat{k}, 3 \hat{i} + 5 \hat{j} + 2 \hat{k}$
original image
$\overset{―}{AB} = 3 \hat{i} - \hat{j} - 2 \hat{k}$
$\overset{―}{BC} = - 2 \hat{i} + 3 \hat{j} - \hat{k}$
$\vec{AC} = \hat{i} + 2 \hat{j} - 3 \hat{k}$
$| \vec{AB} | = \sqrt{9 + 1 + 4} = \sqrt{14}$
$| \vec{BC} | = \sqrt{4 + 9 + 1} = \sqrt{14}$
$| \vec{AC} | = \sqrt{1 + 4 + 9} = \sqrt{14}$
$△ ABC$ is an equilateral triangle
$∴$ Orthocentre of triangle is coincide with centroid
$∴$ Orthocentre of $△ ABC$ is
$(\frac{2 \hat{i} + 3 \hat{j} + 5 \hat{k} + 5 \hat{i} + 2 \hat{j} + 3 \hat{k} + 3 \hat{i} + 5 \hat{j} + 2 \hat{k}}{3})$
$= \frac{15 \hat{i} + 15 \hat{j} + 15 \hat{k}}{3} = 5 \hat{i} + 5 \hat{j} + 5 \hat{k}$
So, $x = y = z = 5$

TS EAMCET-2020-10.09.2020

Vector Algebra

87714 If $P, Q, R$ are the mid-point of the sides $A B, B C$ and $C A$ of $△ A B C$ respectively, then $P C - B Q =$

1

CP

2

PQ

3

BR

4

AR

Explanation:

, d) : Let position vectors of vertices A,B,C are $a, b, c$ respectively
Since, $P, Q, R$ are mid points of $A B, B C, C A$
original image
$∴$ Position vector of $P, Q, R$ are
$(\frac{a + b}{2}) (\frac{b + c}{2}) (\frac{c + a}{2})$
Respectively
Now,
$PC - BQ = [c - (\frac{a + b}{2})] - [(\frac{b + c}{2}) - b]$
$= \frac{2 c - a - b}{2} - \frac{b + c - 2 b}{2} = \frac{2 c - a - b - b - c + 2 b}{2} = \frac{c - a}{2}$
Now,
$CP = \frac{a + b}{2} - c = \frac{a + b + 2 c}{2}$
$PQ = \frac{b + c}{2} - \frac{a + b}{2} = \frac{c - a}{2}$
$BR = \frac{c + a}{2} - b = \frac{c - a - 2 b}{2}$
$AR = \frac{c + a}{2} - a = \frac{c - a}{2}$
$∴ PC - BQ = PQ = AR$

TS EAMCET-2018-07.05.2018

Vector Algebra

87715 The position vector of $a$ point $P$ is $2 \hat{i} + \hat{j} + 3 \hat{k}$ and $a = - \hat{i} - 2 \hat{k}, b = \hat{i} + \hat{j} + 2 \hat{k}$ are two vectors which determine a plane $π$ . The equation of a line through $P$ normal to $b$ and lying on the plane $π$ is

1

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- \hat{i} + 5 \hat{j} - 2 \hat{k})

2

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (\hat{i} + \hat{j} + \hat{k})

3

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- 2 \hat{i} - \hat{j} + 3 \hat{k})

4

r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- 3 \hat{i} + 4 \hat{j} - 5 \hat{k})

Explanation:

(A) : Given,
Position vector of a point $P$ is $(2 \hat{i} + \hat{j} + 3 \hat{k})$
Equation of plane,
$(r - 2 \hat{i} + \hat{j} + 3 \hat{k}) \cdot (a \times b) = 0$
$∴$ Normal of plane $a \times b$
Equation of line through $P (2 \hat{i} + \hat{j} + 3 \hat{k})$ and normal to $b$ and lying on the plane is
$r = (2 \hat{i} + \hat{j} + 3 \hat{k}) + λ {a \times (b \times c)}$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ {(b \cdot a) b - (b \cdot b) a}$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ [(- 5) (\hat{i} + \hat{j} + 2 \hat{k}) - 6 (- \hat{i} - 2 \hat{k})]$
$r = 2 \hat{i} + \hat{j} + 3 \hat{k} + λ (- \hat{i} + 5 \hat{j} - 2 \hat{k})$

TS EAMCET-2020-10.09.2020