QBManager

Vector Algebra

88190 Let \(\vec{a}=3 \hat{i}+2 \hat{j}+x \hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\) for some real \(x\). Then \(|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=r\) is possible if

1 \(0\lt\mathrm{r} \leq \sqrt{\frac{3}{2}}\)

2 \(\sqrt{\frac{3}{2}}\lt\mathrm{r} \leq 3 \sqrt{\frac{3}{2}}\)

3 \(3 \sqrt{\frac{3}{2}}\lt\) r \(\lt5 \sqrt{\frac{3}{2}}\)

4 \(r \geq 5 \sqrt{\frac{3}{2}}\)

JEE Main-2019-08.04.2019

Vector Algebra

88191 Let \(\alpha=(\lambda-2) a+b\) and \(\beta=(4 \lambda-2) a+3 b\) be two given vectors where \(a\) and \(b\) are noncollinear. The value of \(\lambda\) for which vectors \(\alpha\) and \(\beta\) are collinear is

1 4

2 -3

3 3

4 -4

JEE Main-2019-10.01.2019

Vector Algebra

88192 If the foot of the perpendicular from point (4, 3,8 ) on the line \(L_1 \Rightarrow \frac{x-a}{1}=\frac{y-2}{3}=\frac{z-b}{4}, l \neq 0\) is \((3,5,7)\), then the shortest distance between the line \(L_1\) and line \(L_2 \Rightarrow \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\) is equal to

1 \(\frac{1}{2}\)

2 \(\frac{1}{\sqrt{6}}\)

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

4 \(\frac{1}{\sqrt{3}}\)

Explanation:

(B) : Given that,
The foot of the perpendicular from point \((4,3,8)\) on the line
\(\mathrm{L}_1=\frac{|\mathrm{x}-\mathrm{a}|}{1}=\frac{|\mathrm{y}-2|}{3}=\frac{|\mathrm{z}-\mathrm{b}|}{4} \text { is }(3,5,7)\)
\((3,5,7)\) lies on the line \(\mathrm{L}_1\)
Hence, Its co-ordinates satisfy the equation of the line,
\(\frac{3-\mathrm{a}}{l}=\frac{5-2}{3}=\frac{7-\mathrm{b}}{4} \Rightarrow \frac{3-\mathrm{a}}{l}=1=\frac{7-\mathrm{b}}{4}\)
\(\frac{3-\mathrm{a}}{l}=1 \text { and } \frac{7-\mathrm{b}}{4}=1\)
\(\mathrm{a}+\mathrm{l}=3\) and \(\mathrm{b}=3\)
Let, \(A=(4,3,8)\) and \(B=(3,5,7)\)
\(\overrightarrow{\mathrm{AB}}=(4-3) \hat{\mathrm{i}}+(3-5) \hat{\mathrm{j}}+(8-7) \hat{\mathrm{k}}\)
\(\overline{\mathrm{AB}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)
For a line,
\(\frac{\left|\mathrm{x}-\mathrm{x}_1\right|}{\mathrm{a}_1}=\frac{\left|\mathrm{y}-\mathrm{y}_1\right|}{\mathrm{b}_1}=\frac{\left|\mathrm{z}-\mathrm{z}_1\right|}{\mathrm{c}_1}\)
\(\left(a_1, b_1, c_1\right)\) are the direction ratios the direction radios of \(\mathrm{L}_1\) are \((l, 3,4) \overrightarrow{\mathrm{AB}}\) is perpendicular to line \(\mathrm{L}_1\)
\(\overrightarrow{\mathrm{AB}} \cdot \overline{\mathrm{L}_1}=0\)
\(\Rightarrow \quad(\hat{i}-2 \hat{j}+\hat{k}) \cdot(l \hat{i}+3 \hat{j}+4 \hat{k})=0\)
\(1 \times l-2 \times 3+1 \times 4=0\)
\(l=2\)
Now,
From equation (i),
\(\mathrm{a}=1\)
Hence, \(\mathrm{L}_1=\frac{|\mathrm{x}-1|}{2}=\frac{|\mathrm{y}-2|}{3}=\frac{|\mathrm{z}-3|}{4}\) and
\(\mathrm{L}_2=\frac{|\mathrm{x}-2|}{3}=\frac{|\mathrm{y}-4|}{4}=\frac{|\mathrm{z}-5|}{5}\) are the skew lines.
Now,
The shortest distance between two skew lines are
\(\frac{\left|\mathrm{x}-\mathrm{x}_1\right|}{\mathrm{a}_1}=\frac{\left|\mathrm{y}-\mathrm{y}_1\right|}{\mathrm{b}_1}=\frac{\left|\mathrm{z}-\mathrm{z}_1\right|}{\mathrm{c}_1} \text { and }\)
\(\frac{\left|\mathrm{x}-\mathrm{x}_2\right|}{\mathrm{a}_2}=\frac{\left|\mathrm{y}-\mathrm{y}_2\right|}{\mathrm{b}_2}=\frac{\left|\mathrm{z}-\mathrm{z}_2\right|}{\mathrm{c}_2} \text { is given as }\)
\(d=\frac{\left|\begin{array}{ccc}
x_2-x_1& y_2-y_1 &z_2-z_1\\
a_1 &b_1& c_1\\
a_2 &b_2 &c_2
\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}\)
Substituting all the required values we get -
\(\mathrm{d}=\frac{\left|\begin{array}{ccc}
2-1& 4-2 &5-3\\
2 &3& 4\\
3 &4 &5
\end{array}\right|}{\sqrt{(15-16)^2+(12-10)^2+(8-9)^2}}\)
\(\mathrm{~d}=\frac{1(15-16)-2(10-12)+2(8-9)}{\sqrt{1+4+1}}\)
\(d=\frac{-1+4-2}{\sqrt{6}}\)
\(d=\frac{1}{\sqrt{6}} \text { units }\)
Hence, the shortest distance between the lines \(\mathrm{L}_1\) and \(\mathrm{L}_2\) is \(\frac{1}{\sqrt{6}}\) units.

JEE Main-2021-16.03.2021

Vector Algebra

88193 Three concurrent edges of a parallelepiped are given by
\(\mathbf{a}=\mathbf{2} \hat{\mathbf{i}}-\mathbf{3} \hat{\mathbf{i}}+\hat{\mathbf{k}},\)
\(\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\mathbf{2} \hat{\mathbf{k}},\)
\(\mathbf{c}=\mathbf{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\)
and
The volume of the parallelepiped is

1 14 cu units

2 20 cu units

3 \(25 \mathrm{cu}\) units

4 \(60 \mathrm{cu}\) units

Manipal-2020

Vector Algebra

88190 Let \(\vec{a}=3 \hat{i}+2 \hat{j}+x \hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\) for some real \(x\). Then \(|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=r\) is possible if

1 \(0\lt\mathrm{r} \leq \sqrt{\frac{3}{2}}\)

2 \(\sqrt{\frac{3}{2}}\lt\mathrm{r} \leq 3 \sqrt{\frac{3}{2}}\)

3 \(3 \sqrt{\frac{3}{2}}\lt\) r \(\lt5 \sqrt{\frac{3}{2}}\)

4 \(r \geq 5 \sqrt{\frac{3}{2}}\)

JEE Main-2019-08.04.2019

Vector Algebra

88191 Let \(\alpha=(\lambda-2) a+b\) and \(\beta=(4 \lambda-2) a+3 b\) be two given vectors where \(a\) and \(b\) are noncollinear. The value of \(\lambda\) for which vectors \(\alpha\) and \(\beta\) are collinear is

1 4

2 -3

3 3

4 -4

JEE Main-2019-10.01.2019

Vector Algebra

88192 If the foot of the perpendicular from point (4, 3,8 ) on the line \(L_1 \Rightarrow \frac{x-a}{1}=\frac{y-2}{3}=\frac{z-b}{4}, l \neq 0\) is \((3,5,7)\), then the shortest distance between the line \(L_1\) and line \(L_2 \Rightarrow \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\) is equal to

1 \(\frac{1}{2}\)

2 \(\frac{1}{\sqrt{6}}\)

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

4 \(\frac{1}{\sqrt{3}}\)

Explanation:

(B) : Given that,
The foot of the perpendicular from point \((4,3,8)\) on the line
\(\mathrm{L}_1=\frac{|\mathrm{x}-\mathrm{a}|}{1}=\frac{|\mathrm{y}-2|}{3}=\frac{|\mathrm{z}-\mathrm{b}|}{4} \text { is }(3,5,7)\)
\((3,5,7)\) lies on the line \(\mathrm{L}_1\)
Hence, Its co-ordinates satisfy the equation of the line,
\(\frac{3-\mathrm{a}}{l}=\frac{5-2}{3}=\frac{7-\mathrm{b}}{4} \Rightarrow \frac{3-\mathrm{a}}{l}=1=\frac{7-\mathrm{b}}{4}\)
\(\frac{3-\mathrm{a}}{l}=1 \text { and } \frac{7-\mathrm{b}}{4}=1\)
\(\mathrm{a}+\mathrm{l}=3\) and \(\mathrm{b}=3\)
Let, \(A=(4,3,8)\) and \(B=(3,5,7)\)
\(\overrightarrow{\mathrm{AB}}=(4-3) \hat{\mathrm{i}}+(3-5) \hat{\mathrm{j}}+(8-7) \hat{\mathrm{k}}\)
\(\overline{\mathrm{AB}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)
For a line,
\(\frac{\left|\mathrm{x}-\mathrm{x}_1\right|}{\mathrm{a}_1}=\frac{\left|\mathrm{y}-\mathrm{y}_1\right|}{\mathrm{b}_1}=\frac{\left|\mathrm{z}-\mathrm{z}_1\right|}{\mathrm{c}_1}\)
\(\left(a_1, b_1, c_1\right)\) are the direction ratios the direction radios of \(\mathrm{L}_1\) are \((l, 3,4) \overrightarrow{\mathrm{AB}}\) is perpendicular to line \(\mathrm{L}_1\)
\(\overrightarrow{\mathrm{AB}} \cdot \overline{\mathrm{L}_1}=0\)
\(\Rightarrow \quad(\hat{i}-2 \hat{j}+\hat{k}) \cdot(l \hat{i}+3 \hat{j}+4 \hat{k})=0\)
\(1 \times l-2 \times 3+1 \times 4=0\)
\(l=2\)
Now,
From equation (i),
\(\mathrm{a}=1\)
Hence, \(\mathrm{L}_1=\frac{|\mathrm{x}-1|}{2}=\frac{|\mathrm{y}-2|}{3}=\frac{|\mathrm{z}-3|}{4}\) and
\(\mathrm{L}_2=\frac{|\mathrm{x}-2|}{3}=\frac{|\mathrm{y}-4|}{4}=\frac{|\mathrm{z}-5|}{5}\) are the skew lines.
Now,
The shortest distance between two skew lines are
\(\frac{\left|\mathrm{x}-\mathrm{x}_1\right|}{\mathrm{a}_1}=\frac{\left|\mathrm{y}-\mathrm{y}_1\right|}{\mathrm{b}_1}=\frac{\left|\mathrm{z}-\mathrm{z}_1\right|}{\mathrm{c}_1} \text { and }\)
\(\frac{\left|\mathrm{x}-\mathrm{x}_2\right|}{\mathrm{a}_2}=\frac{\left|\mathrm{y}-\mathrm{y}_2\right|}{\mathrm{b}_2}=\frac{\left|\mathrm{z}-\mathrm{z}_2\right|}{\mathrm{c}_2} \text { is given as }\)
\(d=\frac{\left|\begin{array}{ccc}
x_2-x_1& y_2-y_1 &z_2-z_1\\
a_1 &b_1& c_1\\
a_2 &b_2 &c_2
\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}\)
Substituting all the required values we get -
\(\mathrm{d}=\frac{\left|\begin{array}{ccc}
2-1& 4-2 &5-3\\
2 &3& 4\\
3 &4 &5
\end{array}\right|}{\sqrt{(15-16)^2+(12-10)^2+(8-9)^2}}\)
\(\mathrm{~d}=\frac{1(15-16)-2(10-12)+2(8-9)}{\sqrt{1+4+1}}\)
\(d=\frac{-1+4-2}{\sqrt{6}}\)
\(d=\frac{1}{\sqrt{6}} \text { units }\)
Hence, the shortest distance between the lines \(\mathrm{L}_1\) and \(\mathrm{L}_2\) is \(\frac{1}{\sqrt{6}}\) units.

JEE Main-2021-16.03.2021

Vector Algebra

88193 Three concurrent edges of a parallelepiped are given by
\(\mathbf{a}=\mathbf{2} \hat{\mathbf{i}}-\mathbf{3} \hat{\mathbf{i}}+\hat{\mathbf{k}},\)
\(\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\mathbf{2} \hat{\mathbf{k}},\)
\(\mathbf{c}=\mathbf{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\)
and
The volume of the parallelepiped is

1 14 cu units

2 20 cu units

3 \(25 \mathrm{cu}\) units

4 \(60 \mathrm{cu}\) units

Manipal-2020

Vector Algebra

88190 Let \(\vec{a}=3 \hat{i}+2 \hat{j}+x \hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\) for some real \(x\). Then \(|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=r\) is possible if

1 \(0\lt\mathrm{r} \leq \sqrt{\frac{3}{2}}\)

2 \(\sqrt{\frac{3}{2}}\lt\mathrm{r} \leq 3 \sqrt{\frac{3}{2}}\)

3 \(3 \sqrt{\frac{3}{2}}\lt\) r \(\lt5 \sqrt{\frac{3}{2}}\)

4 \(r \geq 5 \sqrt{\frac{3}{2}}\)

JEE Main-2019-08.04.2019

Vector Algebra

88191 Let \(\alpha=(\lambda-2) a+b\) and \(\beta=(4 \lambda-2) a+3 b\) be two given vectors where \(a\) and \(b\) are noncollinear. The value of \(\lambda\) for which vectors \(\alpha\) and \(\beta\) are collinear is

1 4

2 -3

3 3

4 -4

JEE Main-2019-10.01.2019

Vector Algebra

88192 If the foot of the perpendicular from point (4, 3,8 ) on the line \(L_1 \Rightarrow \frac{x-a}{1}=\frac{y-2}{3}=\frac{z-b}{4}, l \neq 0\) is \((3,5,7)\), then the shortest distance between the line \(L_1\) and line \(L_2 \Rightarrow \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\) is equal to

1 \(\frac{1}{2}\)

2 \(\frac{1}{\sqrt{6}}\)

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

4 \(\frac{1}{\sqrt{3}}\)

Explanation:

(B) : Given that,
The foot of the perpendicular from point \((4,3,8)\) on the line
\(\mathrm{L}_1=\frac{|\mathrm{x}-\mathrm{a}|}{1}=\frac{|\mathrm{y}-2|}{3}=\frac{|\mathrm{z}-\mathrm{b}|}{4} \text { is }(3,5,7)\)
\((3,5,7)\) lies on the line \(\mathrm{L}_1\)
Hence, Its co-ordinates satisfy the equation of the line,
\(\frac{3-\mathrm{a}}{l}=\frac{5-2}{3}=\frac{7-\mathrm{b}}{4} \Rightarrow \frac{3-\mathrm{a}}{l}=1=\frac{7-\mathrm{b}}{4}\)
\(\frac{3-\mathrm{a}}{l}=1 \text { and } \frac{7-\mathrm{b}}{4}=1\)
\(\mathrm{a}+\mathrm{l}=3\) and \(\mathrm{b}=3\)
Let, \(A=(4,3,8)\) and \(B=(3,5,7)\)
\(\overrightarrow{\mathrm{AB}}=(4-3) \hat{\mathrm{i}}+(3-5) \hat{\mathrm{j}}+(8-7) \hat{\mathrm{k}}\)
\(\overline{\mathrm{AB}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)
For a line,
\(\frac{\left|\mathrm{x}-\mathrm{x}_1\right|}{\mathrm{a}_1}=\frac{\left|\mathrm{y}-\mathrm{y}_1\right|}{\mathrm{b}_1}=\frac{\left|\mathrm{z}-\mathrm{z}_1\right|}{\mathrm{c}_1}\)
\(\left(a_1, b_1, c_1\right)\) are the direction ratios the direction radios of \(\mathrm{L}_1\) are \((l, 3,4) \overrightarrow{\mathrm{AB}}\) is perpendicular to line \(\mathrm{L}_1\)
\(\overrightarrow{\mathrm{AB}} \cdot \overline{\mathrm{L}_1}=0\)
\(\Rightarrow \quad(\hat{i}-2 \hat{j}+\hat{k}) \cdot(l \hat{i}+3 \hat{j}+4 \hat{k})=0\)
\(1 \times l-2 \times 3+1 \times 4=0\)
\(l=2\)
Now,
From equation (i),
\(\mathrm{a}=1\)
Hence, \(\mathrm{L}_1=\frac{|\mathrm{x}-1|}{2}=\frac{|\mathrm{y}-2|}{3}=\frac{|\mathrm{z}-3|}{4}\) and
\(\mathrm{L}_2=\frac{|\mathrm{x}-2|}{3}=\frac{|\mathrm{y}-4|}{4}=\frac{|\mathrm{z}-5|}{5}\) are the skew lines.
Now,
The shortest distance between two skew lines are
\(\frac{\left|\mathrm{x}-\mathrm{x}_1\right|}{\mathrm{a}_1}=\frac{\left|\mathrm{y}-\mathrm{y}_1\right|}{\mathrm{b}_1}=\frac{\left|\mathrm{z}-\mathrm{z}_1\right|}{\mathrm{c}_1} \text { and }\)
\(\frac{\left|\mathrm{x}-\mathrm{x}_2\right|}{\mathrm{a}_2}=\frac{\left|\mathrm{y}-\mathrm{y}_2\right|}{\mathrm{b}_2}=\frac{\left|\mathrm{z}-\mathrm{z}_2\right|}{\mathrm{c}_2} \text { is given as }\)
\(d=\frac{\left|\begin{array}{ccc}
x_2-x_1& y_2-y_1 &z_2-z_1\\
a_1 &b_1& c_1\\
a_2 &b_2 &c_2
\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}\)
Substituting all the required values we get -
\(\mathrm{d}=\frac{\left|\begin{array}{ccc}
2-1& 4-2 &5-3\\
2 &3& 4\\
3 &4 &5
\end{array}\right|}{\sqrt{(15-16)^2+(12-10)^2+(8-9)^2}}\)
\(\mathrm{~d}=\frac{1(15-16)-2(10-12)+2(8-9)}{\sqrt{1+4+1}}\)
\(d=\frac{-1+4-2}{\sqrt{6}}\)
\(d=\frac{1}{\sqrt{6}} \text { units }\)
Hence, the shortest distance between the lines \(\mathrm{L}_1\) and \(\mathrm{L}_2\) is \(\frac{1}{\sqrt{6}}\) units.

JEE Main-2021-16.03.2021

Vector Algebra

88193 Three concurrent edges of a parallelepiped are given by
\(\mathbf{a}=\mathbf{2} \hat{\mathbf{i}}-\mathbf{3} \hat{\mathbf{i}}+\hat{\mathbf{k}},\)
\(\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\mathbf{2} \hat{\mathbf{k}},\)
\(\mathbf{c}=\mathbf{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\)
and
The volume of the parallelepiped is

1 14 cu units

2 20 cu units

3 \(25 \mathrm{cu}\) units

4 \(60 \mathrm{cu}\) units

Manipal-2020

Vector Algebra

88190 Let \(\vec{a}=3 \hat{i}+2 \hat{j}+x \hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\) for some real \(x\). Then \(|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=r\) is possible if

1 \(0\lt\mathrm{r} \leq \sqrt{\frac{3}{2}}\)

2 \(\sqrt{\frac{3}{2}}\lt\mathrm{r} \leq 3 \sqrt{\frac{3}{2}}\)

3 \(3 \sqrt{\frac{3}{2}}\lt\) r \(\lt5 \sqrt{\frac{3}{2}}\)

4 \(r \geq 5 \sqrt{\frac{3}{2}}\)

JEE Main-2019-08.04.2019

Vector Algebra

88191 Let \(\alpha=(\lambda-2) a+b\) and \(\beta=(4 \lambda-2) a+3 b\) be two given vectors where \(a\) and \(b\) are noncollinear. The value of \(\lambda\) for which vectors \(\alpha\) and \(\beta\) are collinear is

1 4

2 -3

3 3

4 -4

JEE Main-2019-10.01.2019

Vector Algebra

88192 If the foot of the perpendicular from point (4, 3,8 ) on the line \(L_1 \Rightarrow \frac{x-a}{1}=\frac{y-2}{3}=\frac{z-b}{4}, l \neq 0\) is \((3,5,7)\), then the shortest distance between the line \(L_1\) and line \(L_2 \Rightarrow \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\) is equal to

1 \(\frac{1}{2}\)

2 \(\frac{1}{\sqrt{6}}\)

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

4 \(\frac{1}{\sqrt{3}}\)

Explanation:

(B) : Given that,
The foot of the perpendicular from point \((4,3,8)\) on the line
\(\mathrm{L}_1=\frac{|\mathrm{x}-\mathrm{a}|}{1}=\frac{|\mathrm{y}-2|}{3}=\frac{|\mathrm{z}-\mathrm{b}|}{4} \text { is }(3,5,7)\)
\((3,5,7)\) lies on the line \(\mathrm{L}_1\)
Hence, Its co-ordinates satisfy the equation of the line,
\(\frac{3-\mathrm{a}}{l}=\frac{5-2}{3}=\frac{7-\mathrm{b}}{4} \Rightarrow \frac{3-\mathrm{a}}{l}=1=\frac{7-\mathrm{b}}{4}\)
\(\frac{3-\mathrm{a}}{l}=1 \text { and } \frac{7-\mathrm{b}}{4}=1\)
\(\mathrm{a}+\mathrm{l}=3\) and \(\mathrm{b}=3\)
Let, \(A=(4,3,8)\) and \(B=(3,5,7)\)
\(\overrightarrow{\mathrm{AB}}=(4-3) \hat{\mathrm{i}}+(3-5) \hat{\mathrm{j}}+(8-7) \hat{\mathrm{k}}\)
\(\overline{\mathrm{AB}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)
For a line,
\(\frac{\left|\mathrm{x}-\mathrm{x}_1\right|}{\mathrm{a}_1}=\frac{\left|\mathrm{y}-\mathrm{y}_1\right|}{\mathrm{b}_1}=\frac{\left|\mathrm{z}-\mathrm{z}_1\right|}{\mathrm{c}_1}\)
\(\left(a_1, b_1, c_1\right)\) are the direction ratios the direction radios of \(\mathrm{L}_1\) are \((l, 3,4) \overrightarrow{\mathrm{AB}}\) is perpendicular to line \(\mathrm{L}_1\)
\(\overrightarrow{\mathrm{AB}} \cdot \overline{\mathrm{L}_1}=0\)
\(\Rightarrow \quad(\hat{i}-2 \hat{j}+\hat{k}) \cdot(l \hat{i}+3 \hat{j}+4 \hat{k})=0\)
\(1 \times l-2 \times 3+1 \times 4=0\)
\(l=2\)
Now,
From equation (i),
\(\mathrm{a}=1\)
Hence, \(\mathrm{L}_1=\frac{|\mathrm{x}-1|}{2}=\frac{|\mathrm{y}-2|}{3}=\frac{|\mathrm{z}-3|}{4}\) and
\(\mathrm{L}_2=\frac{|\mathrm{x}-2|}{3}=\frac{|\mathrm{y}-4|}{4}=\frac{|\mathrm{z}-5|}{5}\) are the skew lines.
Now,
The shortest distance between two skew lines are
\(\frac{\left|\mathrm{x}-\mathrm{x}_1\right|}{\mathrm{a}_1}=\frac{\left|\mathrm{y}-\mathrm{y}_1\right|}{\mathrm{b}_1}=\frac{\left|\mathrm{z}-\mathrm{z}_1\right|}{\mathrm{c}_1} \text { and }\)
\(\frac{\left|\mathrm{x}-\mathrm{x}_2\right|}{\mathrm{a}_2}=\frac{\left|\mathrm{y}-\mathrm{y}_2\right|}{\mathrm{b}_2}=\frac{\left|\mathrm{z}-\mathrm{z}_2\right|}{\mathrm{c}_2} \text { is given as }\)
\(d=\frac{\left|\begin{array}{ccc}
x_2-x_1& y_2-y_1 &z_2-z_1\\
a_1 &b_1& c_1\\
a_2 &b_2 &c_2
\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}\)
Substituting all the required values we get -
\(\mathrm{d}=\frac{\left|\begin{array}{ccc}
2-1& 4-2 &5-3\\
2 &3& 4\\
3 &4 &5
\end{array}\right|}{\sqrt{(15-16)^2+(12-10)^2+(8-9)^2}}\)
\(\mathrm{~d}=\frac{1(15-16)-2(10-12)+2(8-9)}{\sqrt{1+4+1}}\)
\(d=\frac{-1+4-2}{\sqrt{6}}\)
\(d=\frac{1}{\sqrt{6}} \text { units }\)
Hence, the shortest distance between the lines \(\mathrm{L}_1\) and \(\mathrm{L}_2\) is \(\frac{1}{\sqrt{6}}\) units.

JEE Main-2021-16.03.2021

Vector Algebra

88193 Three concurrent edges of a parallelepiped are given by
\(\mathbf{a}=\mathbf{2} \hat{\mathbf{i}}-\mathbf{3} \hat{\mathbf{i}}+\hat{\mathbf{k}},\)
\(\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\mathbf{2} \hat{\mathbf{k}},\)
\(\mathbf{c}=\mathbf{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\)
and
The volume of the parallelepiped is

1 14 cu units

2 20 cu units

3 \(25 \mathrm{cu}\) units

4 \(60 \mathrm{cu}\) units

Manipal-2020