QBManager

Three Dimensional Geometry

121217 Two lines
\(\mathrm{L}_1: \mathrm{x}=5, \frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}, \mathrm{~L}_2: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}\) are coplanar. Then \(\alpha\) can take value (s)

1 \(1,2,5\)

2 \(1,4,5\)

3 \(3,4,5\)

4 \(2,4,5\)

Explanation:

A Two lines are given as -
\(\mathrm{L}_1: \mathrm{x}=5, \frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}\) or \(\frac{\mathrm{x}-5}{1}=\frac{\mathrm{y}-0}{3-\alpha}=\frac{\mathrm{z}-0}{-2}\)
\(\mathrm{L}_2: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}\) or \(\frac{\mathrm{x}-\alpha}{1}=\frac{\mathrm{y}-0}{-1}=\frac{\mathrm{z}-0}{2-\alpha}\)
We know that,
For line, \(\mathrm{L}_1 \frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{a}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~b}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{c}_1}\)
\(\frac{x-5}{1}=\frac{y-0}{3-\alpha}=\frac{z-0}{-2}\) Comparing by equation (i) \(\mathrm{x}_1=5, \mathrm{y}_1=0, \mathrm{z}_1=0\)
\(\mathrm{a}_1=1, \mathrm{~b}_1=3-\alpha, \mathrm{c}_1=-2\)
For line \(L_2, \frac{x-\alpha}{1}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha}\)
\(\mathrm{x}_2=+\alpha, \mathrm{y}_2=0, \mathrm{z}_2=0\) comparing by equation (ii),
\(\mathrm{a}_2=1, \mathrm{~b}_2=-1, \mathrm{c}_2=2-\alpha\)
Two lines are coplanar-
\(\Rightarrow \quad\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|=0\)
\(\Rightarrow \quad\left|\begin{array}{lcc}\alpha-5 & 0-0 & 0-0 \\ 1 & 3-\alpha & -2 \\ 1 & -1 & 2-\alpha\end{array}\right|=0\)
\(\Rightarrow \quad\left|\begin{array}{ccc}\alpha-5 & 0 & 0 \\ 1 & 3-\alpha & -2 \\ 1 & -1 & 2-\alpha\end{array}\right|=0\)
\(\Rightarrow \quad(\alpha-5)[(3-\alpha)(2-\alpha)-2]=0\)
\(\Rightarrow \quad(\alpha-5)\left[6-3 \alpha-2 \alpha+\alpha^2-2\right]=0\)
\(\Rightarrow \quad(\alpha-5)\left(\alpha^2-5 \alpha+4\right)=0\)
\(\Rightarrow \quad(\alpha-5)\left(\alpha^2-4 \alpha-\alpha+4\right)=0\)
\(\Rightarrow \quad(\alpha-5)[\alpha(\alpha-4)-1(\alpha-4)]=0\)
\(\Rightarrow \quad(\alpha-5)(\alpha-4)(\alpha-1)=0\)
So,\(\alpha=1,4,5\)

BITSAT-2015

Three Dimensional Geometry

121220 If the origin and the points \((1,2,3),(2,3,4)\) and \((x, y, z)\) are coplanar, then

1 \(z-2 x+y=0\)

2 \(x-2 y+z+1=0\)

3 \(x+y+z=6\)

4 \(x-2 y+z=0\)

MHT CET-2020

Three Dimensional Geometry

121221 The shortest distance between the lines \(\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}\) and \(\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}\) is

1 \(7 \sqrt{3}\)

2 \(6 \sqrt{3}\)

3 \(4 \sqrt{3}\)

4 \(5 \sqrt{3}\)

JEE Main-01.02.2023

Three Dimensional Geometry

121223 If the shortest distance between the lines
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda} \text { and } \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}\)
is \(\frac{1}{\sqrt{3}}\), then the sum of all possible values of \(\lambda\) is:

1 16

2 6

3 12

4 15

Explanation:

A Given that,
The lines, \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}\)
And, \(\quad \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}\)
On comparing the equations (i) and (ii), we get -
\(\overrightarrow{\mathrm{a}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{a}}_2=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{M}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{N}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
\(\therefore \quad \overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & \lambda \\ 1 & 4 & 5\end{array}\right|\)
\(=\hat{\mathrm{i}}(15-4 \lambda)-\hat{\mathrm{j}}(10-\lambda)+\hat{\mathrm{k}}(8-3)\)
\(=(15-4 \lambda) \hat{\mathrm{i}}-(10-\lambda) \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
And, \(\overrightarrow{\mathrm{a}}_2-\overrightarrow{\mathrm{a}}_1=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})-(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\)
\(=\hat{i}+2 \hat{j}+2 \hat{k}\)
So, shortest distance, \(\mathrm{d}\)
\(=\left|\frac{\left(\mathrm{a}_2-\mathrm{a}_1\right) \cdot(\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}})}{|\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}}|}\right|\)
\(\Rightarrow\left|\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot[(15-4 \lambda) \hat{i}-(10-\lambda) \hat{j}+5 \hat{k}]}{\sqrt{(15-4 \lambda)^2+(10-\lambda)^2+5^2}}\right|=\frac{1}{\sqrt{3}}\)
\(\Rightarrow\left|\frac{(15-4 \lambda)-2(10-\lambda)+10}{\sqrt{(15-4 \lambda)^2+(10-\lambda)^2+5^2}}\right|=\frac{1}{\sqrt{3}}\)
On squaring both sides,
\(\frac{(5-2 \lambda)^2}{(15-4 \lambda)^2+(10-\lambda)^2+25}=\frac{1}{3}\)
\(\Rightarrow 3(5-2 \lambda)^2=(15-4 \lambda)^2+(10-\lambda)^2+25\)
By solving the equation we get -
\(\Rightarrow 5 \lambda^2-80 \lambda+275=0\)
\(\Rightarrow \lambda^2-16 \lambda+55=0\)
\(\Rightarrow(\lambda-5)(\lambda-11)=0\)
\(\Rightarrow \lambda=5,11\)
Hence,
The sum of all possible value of \(\lambda\),
\(\lambda=5+11\)
\(\therefore \quad \lambda =16\)

JEE Main-24.06.2022

Three Dimensional Geometry

121217 Two lines
\(\mathrm{L}_1: \mathrm{x}=5, \frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}, \mathrm{~L}_2: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}\) are coplanar. Then \(\alpha\) can take value (s)

1 \(1,2,5\)

2 \(1,4,5\)

3 \(3,4,5\)

4 \(2,4,5\)

Explanation:

A Two lines are given as -
\(\mathrm{L}_1: \mathrm{x}=5, \frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}\) or \(\frac{\mathrm{x}-5}{1}=\frac{\mathrm{y}-0}{3-\alpha}=\frac{\mathrm{z}-0}{-2}\)
\(\mathrm{L}_2: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}\) or \(\frac{\mathrm{x}-\alpha}{1}=\frac{\mathrm{y}-0}{-1}=\frac{\mathrm{z}-0}{2-\alpha}\)
We know that,
For line, \(\mathrm{L}_1 \frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{a}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~b}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{c}_1}\)
\(\frac{x-5}{1}=\frac{y-0}{3-\alpha}=\frac{z-0}{-2}\) Comparing by equation (i) \(\mathrm{x}_1=5, \mathrm{y}_1=0, \mathrm{z}_1=0\)
\(\mathrm{a}_1=1, \mathrm{~b}_1=3-\alpha, \mathrm{c}_1=-2\)
For line \(L_2, \frac{x-\alpha}{1}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha}\)
\(\mathrm{x}_2=+\alpha, \mathrm{y}_2=0, \mathrm{z}_2=0\) comparing by equation (ii),
\(\mathrm{a}_2=1, \mathrm{~b}_2=-1, \mathrm{c}_2=2-\alpha\)
Two lines are coplanar-
\(\Rightarrow \quad\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|=0\)
\(\Rightarrow \quad\left|\begin{array}{lcc}\alpha-5 & 0-0 & 0-0 \\ 1 & 3-\alpha & -2 \\ 1 & -1 & 2-\alpha\end{array}\right|=0\)
\(\Rightarrow \quad\left|\begin{array}{ccc}\alpha-5 & 0 & 0 \\ 1 & 3-\alpha & -2 \\ 1 & -1 & 2-\alpha\end{array}\right|=0\)
\(\Rightarrow \quad(\alpha-5)[(3-\alpha)(2-\alpha)-2]=0\)
\(\Rightarrow \quad(\alpha-5)\left[6-3 \alpha-2 \alpha+\alpha^2-2\right]=0\)
\(\Rightarrow \quad(\alpha-5)\left(\alpha^2-5 \alpha+4\right)=0\)
\(\Rightarrow \quad(\alpha-5)\left(\alpha^2-4 \alpha-\alpha+4\right)=0\)
\(\Rightarrow \quad(\alpha-5)[\alpha(\alpha-4)-1(\alpha-4)]=0\)
\(\Rightarrow \quad(\alpha-5)(\alpha-4)(\alpha-1)=0\)
So,\(\alpha=1,4,5\)

BITSAT-2015

Three Dimensional Geometry

121220 If the origin and the points \((1,2,3),(2,3,4)\) and \((x, y, z)\) are coplanar, then

1 \(z-2 x+y=0\)

2 \(x-2 y+z+1=0\)

3 \(x+y+z=6\)

4 \(x-2 y+z=0\)

MHT CET-2020

Three Dimensional Geometry

121221 The shortest distance between the lines \(\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}\) and \(\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}\) is

1 \(7 \sqrt{3}\)

2 \(6 \sqrt{3}\)

3 \(4 \sqrt{3}\)

4 \(5 \sqrt{3}\)

JEE Main-01.02.2023

Three Dimensional Geometry

121223 If the shortest distance between the lines
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda} \text { and } \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}\)
is \(\frac{1}{\sqrt{3}}\), then the sum of all possible values of \(\lambda\) is:

1 16

2 6

3 12

4 15

Explanation:

A Given that,
The lines, \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}\)
And, \(\quad \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}\)
On comparing the equations (i) and (ii), we get -
\(\overrightarrow{\mathrm{a}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{a}}_2=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{M}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{N}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
\(\therefore \quad \overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & \lambda \\ 1 & 4 & 5\end{array}\right|\)
\(=\hat{\mathrm{i}}(15-4 \lambda)-\hat{\mathrm{j}}(10-\lambda)+\hat{\mathrm{k}}(8-3)\)
\(=(15-4 \lambda) \hat{\mathrm{i}}-(10-\lambda) \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
And, \(\overrightarrow{\mathrm{a}}_2-\overrightarrow{\mathrm{a}}_1=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})-(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\)
\(=\hat{i}+2 \hat{j}+2 \hat{k}\)
So, shortest distance, \(\mathrm{d}\)
\(=\left|\frac{\left(\mathrm{a}_2-\mathrm{a}_1\right) \cdot(\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}})}{|\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}}|}\right|\)
\(\Rightarrow\left|\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot[(15-4 \lambda) \hat{i}-(10-\lambda) \hat{j}+5 \hat{k}]}{\sqrt{(15-4 \lambda)^2+(10-\lambda)^2+5^2}}\right|=\frac{1}{\sqrt{3}}\)
\(\Rightarrow\left|\frac{(15-4 \lambda)-2(10-\lambda)+10}{\sqrt{(15-4 \lambda)^2+(10-\lambda)^2+5^2}}\right|=\frac{1}{\sqrt{3}}\)
On squaring both sides,
\(\frac{(5-2 \lambda)^2}{(15-4 \lambda)^2+(10-\lambda)^2+25}=\frac{1}{3}\)
\(\Rightarrow 3(5-2 \lambda)^2=(15-4 \lambda)^2+(10-\lambda)^2+25\)
By solving the equation we get -
\(\Rightarrow 5 \lambda^2-80 \lambda+275=0\)
\(\Rightarrow \lambda^2-16 \lambda+55=0\)
\(\Rightarrow(\lambda-5)(\lambda-11)=0\)
\(\Rightarrow \lambda=5,11\)
Hence,
The sum of all possible value of \(\lambda\),
\(\lambda=5+11\)
\(\therefore \quad \lambda =16\)

JEE Main-24.06.2022

Three Dimensional Geometry

121217 Two lines
\(\mathrm{L}_1: \mathrm{x}=5, \frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}, \mathrm{~L}_2: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}\) are coplanar. Then \(\alpha\) can take value (s)

1 \(1,2,5\)

2 \(1,4,5\)

3 \(3,4,5\)

4 \(2,4,5\)

Explanation:

A Two lines are given as -
\(\mathrm{L}_1: \mathrm{x}=5, \frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}\) or \(\frac{\mathrm{x}-5}{1}=\frac{\mathrm{y}-0}{3-\alpha}=\frac{\mathrm{z}-0}{-2}\)
\(\mathrm{L}_2: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}\) or \(\frac{\mathrm{x}-\alpha}{1}=\frac{\mathrm{y}-0}{-1}=\frac{\mathrm{z}-0}{2-\alpha}\)
We know that,
For line, \(\mathrm{L}_1 \frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{a}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~b}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{c}_1}\)
\(\frac{x-5}{1}=\frac{y-0}{3-\alpha}=\frac{z-0}{-2}\) Comparing by equation (i) \(\mathrm{x}_1=5, \mathrm{y}_1=0, \mathrm{z}_1=0\)
\(\mathrm{a}_1=1, \mathrm{~b}_1=3-\alpha, \mathrm{c}_1=-2\)
For line \(L_2, \frac{x-\alpha}{1}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha}\)
\(\mathrm{x}_2=+\alpha, \mathrm{y}_2=0, \mathrm{z}_2=0\) comparing by equation (ii),
\(\mathrm{a}_2=1, \mathrm{~b}_2=-1, \mathrm{c}_2=2-\alpha\)
Two lines are coplanar-
\(\Rightarrow \quad\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|=0\)
\(\Rightarrow \quad\left|\begin{array}{lcc}\alpha-5 & 0-0 & 0-0 \\ 1 & 3-\alpha & -2 \\ 1 & -1 & 2-\alpha\end{array}\right|=0\)
\(\Rightarrow \quad\left|\begin{array}{ccc}\alpha-5 & 0 & 0 \\ 1 & 3-\alpha & -2 \\ 1 & -1 & 2-\alpha\end{array}\right|=0\)
\(\Rightarrow \quad(\alpha-5)[(3-\alpha)(2-\alpha)-2]=0\)
\(\Rightarrow \quad(\alpha-5)\left[6-3 \alpha-2 \alpha+\alpha^2-2\right]=0\)
\(\Rightarrow \quad(\alpha-5)\left(\alpha^2-5 \alpha+4\right)=0\)
\(\Rightarrow \quad(\alpha-5)\left(\alpha^2-4 \alpha-\alpha+4\right)=0\)
\(\Rightarrow \quad(\alpha-5)[\alpha(\alpha-4)-1(\alpha-4)]=0\)
\(\Rightarrow \quad(\alpha-5)(\alpha-4)(\alpha-1)=0\)
So,\(\alpha=1,4,5\)

BITSAT-2015

Three Dimensional Geometry

121220 If the origin and the points \((1,2,3),(2,3,4)\) and \((x, y, z)\) are coplanar, then

1 \(z-2 x+y=0\)

2 \(x-2 y+z+1=0\)

3 \(x+y+z=6\)

4 \(x-2 y+z=0\)

MHT CET-2020

Three Dimensional Geometry

121221 The shortest distance between the lines \(\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}\) and \(\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}\) is

1 \(7 \sqrt{3}\)

2 \(6 \sqrt{3}\)

3 \(4 \sqrt{3}\)

4 \(5 \sqrt{3}\)

JEE Main-01.02.2023

Three Dimensional Geometry

121223 If the shortest distance between the lines
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda} \text { and } \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}\)
is \(\frac{1}{\sqrt{3}}\), then the sum of all possible values of \(\lambda\) is:

1 16

2 6

3 12

4 15

Explanation:

A Given that,
The lines, \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}\)
And, \(\quad \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}\)
On comparing the equations (i) and (ii), we get -
\(\overrightarrow{\mathrm{a}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{a}}_2=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{M}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{N}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
\(\therefore \quad \overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & \lambda \\ 1 & 4 & 5\end{array}\right|\)
\(=\hat{\mathrm{i}}(15-4 \lambda)-\hat{\mathrm{j}}(10-\lambda)+\hat{\mathrm{k}}(8-3)\)
\(=(15-4 \lambda) \hat{\mathrm{i}}-(10-\lambda) \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
And, \(\overrightarrow{\mathrm{a}}_2-\overrightarrow{\mathrm{a}}_1=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})-(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\)
\(=\hat{i}+2 \hat{j}+2 \hat{k}\)
So, shortest distance, \(\mathrm{d}\)
\(=\left|\frac{\left(\mathrm{a}_2-\mathrm{a}_1\right) \cdot(\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}})}{|\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}}|}\right|\)
\(\Rightarrow\left|\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot[(15-4 \lambda) \hat{i}-(10-\lambda) \hat{j}+5 \hat{k}]}{\sqrt{(15-4 \lambda)^2+(10-\lambda)^2+5^2}}\right|=\frac{1}{\sqrt{3}}\)
\(\Rightarrow\left|\frac{(15-4 \lambda)-2(10-\lambda)+10}{\sqrt{(15-4 \lambda)^2+(10-\lambda)^2+5^2}}\right|=\frac{1}{\sqrt{3}}\)
On squaring both sides,
\(\frac{(5-2 \lambda)^2}{(15-4 \lambda)^2+(10-\lambda)^2+25}=\frac{1}{3}\)
\(\Rightarrow 3(5-2 \lambda)^2=(15-4 \lambda)^2+(10-\lambda)^2+25\)
By solving the equation we get -
\(\Rightarrow 5 \lambda^2-80 \lambda+275=0\)
\(\Rightarrow \lambda^2-16 \lambda+55=0\)
\(\Rightarrow(\lambda-5)(\lambda-11)=0\)
\(\Rightarrow \lambda=5,11\)
Hence,
The sum of all possible value of \(\lambda\),
\(\lambda=5+11\)
\(\therefore \quad \lambda =16\)

JEE Main-24.06.2022

Three Dimensional Geometry

121217 Two lines
\(\mathrm{L}_1: \mathrm{x}=5, \frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}, \mathrm{~L}_2: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}\) are coplanar. Then \(\alpha\) can take value (s)

1 \(1,2,5\)

2 \(1,4,5\)

3 \(3,4,5\)

4 \(2,4,5\)

Explanation:

A Two lines are given as -
\(\mathrm{L}_1: \mathrm{x}=5, \frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}\) or \(\frac{\mathrm{x}-5}{1}=\frac{\mathrm{y}-0}{3-\alpha}=\frac{\mathrm{z}-0}{-2}\)
\(\mathrm{L}_2: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}\) or \(\frac{\mathrm{x}-\alpha}{1}=\frac{\mathrm{y}-0}{-1}=\frac{\mathrm{z}-0}{2-\alpha}\)
We know that,
For line, \(\mathrm{L}_1 \frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{a}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~b}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{c}_1}\)
\(\frac{x-5}{1}=\frac{y-0}{3-\alpha}=\frac{z-0}{-2}\) Comparing by equation (i) \(\mathrm{x}_1=5, \mathrm{y}_1=0, \mathrm{z}_1=0\)
\(\mathrm{a}_1=1, \mathrm{~b}_1=3-\alpha, \mathrm{c}_1=-2\)
For line \(L_2, \frac{x-\alpha}{1}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha}\)
\(\mathrm{x}_2=+\alpha, \mathrm{y}_2=0, \mathrm{z}_2=0\) comparing by equation (ii),
\(\mathrm{a}_2=1, \mathrm{~b}_2=-1, \mathrm{c}_2=2-\alpha\)
Two lines are coplanar-
\(\Rightarrow \quad\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|=0\)
\(\Rightarrow \quad\left|\begin{array}{lcc}\alpha-5 & 0-0 & 0-0 \\ 1 & 3-\alpha & -2 \\ 1 & -1 & 2-\alpha\end{array}\right|=0\)
\(\Rightarrow \quad\left|\begin{array}{ccc}\alpha-5 & 0 & 0 \\ 1 & 3-\alpha & -2 \\ 1 & -1 & 2-\alpha\end{array}\right|=0\)
\(\Rightarrow \quad(\alpha-5)[(3-\alpha)(2-\alpha)-2]=0\)
\(\Rightarrow \quad(\alpha-5)\left[6-3 \alpha-2 \alpha+\alpha^2-2\right]=0\)
\(\Rightarrow \quad(\alpha-5)\left(\alpha^2-5 \alpha+4\right)=0\)
\(\Rightarrow \quad(\alpha-5)\left(\alpha^2-4 \alpha-\alpha+4\right)=0\)
\(\Rightarrow \quad(\alpha-5)[\alpha(\alpha-4)-1(\alpha-4)]=0\)
\(\Rightarrow \quad(\alpha-5)(\alpha-4)(\alpha-1)=0\)
So,\(\alpha=1,4,5\)

BITSAT-2015

Three Dimensional Geometry

121220 If the origin and the points \((1,2,3),(2,3,4)\) and \((x, y, z)\) are coplanar, then

1 \(z-2 x+y=0\)

2 \(x-2 y+z+1=0\)

3 \(x+y+z=6\)

4 \(x-2 y+z=0\)

MHT CET-2020

Three Dimensional Geometry

121221 The shortest distance between the lines \(\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}\) and \(\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}\) is

1 \(7 \sqrt{3}\)

2 \(6 \sqrt{3}\)

3 \(4 \sqrt{3}\)

4 \(5 \sqrt{3}\)

JEE Main-01.02.2023

Three Dimensional Geometry

121223 If the shortest distance between the lines
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda} \text { and } \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}\)
is \(\frac{1}{\sqrt{3}}\), then the sum of all possible values of \(\lambda\) is:

1 16

2 6

3 12

4 15

Explanation:

A Given that,
The lines, \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}\)
And, \(\quad \frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}\)
On comparing the equations (i) and (ii), we get -
\(\overrightarrow{\mathrm{a}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{a}}_2=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{M}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{N}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
\(\therefore \quad \overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & \lambda \\ 1 & 4 & 5\end{array}\right|\)
\(=\hat{\mathrm{i}}(15-4 \lambda)-\hat{\mathrm{j}}(10-\lambda)+\hat{\mathrm{k}}(8-3)\)
\(=(15-4 \lambda) \hat{\mathrm{i}}-(10-\lambda) \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\)
And, \(\overrightarrow{\mathrm{a}}_2-\overrightarrow{\mathrm{a}}_1=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})-(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\)
\(=\hat{i}+2 \hat{j}+2 \hat{k}\)
So, shortest distance, \(\mathrm{d}\)
\(=\left|\frac{\left(\mathrm{a}_2-\mathrm{a}_1\right) \cdot(\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}})}{|\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{N}}|}\right|\)
\(\Rightarrow\left|\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot[(15-4 \lambda) \hat{i}-(10-\lambda) \hat{j}+5 \hat{k}]}{\sqrt{(15-4 \lambda)^2+(10-\lambda)^2+5^2}}\right|=\frac{1}{\sqrt{3}}\)
\(\Rightarrow\left|\frac{(15-4 \lambda)-2(10-\lambda)+10}{\sqrt{(15-4 \lambda)^2+(10-\lambda)^2+5^2}}\right|=\frac{1}{\sqrt{3}}\)
On squaring both sides,
\(\frac{(5-2 \lambda)^2}{(15-4 \lambda)^2+(10-\lambda)^2+25}=\frac{1}{3}\)
\(\Rightarrow 3(5-2 \lambda)^2=(15-4 \lambda)^2+(10-\lambda)^2+25\)
By solving the equation we get -
\(\Rightarrow 5 \lambda^2-80 \lambda+275=0\)
\(\Rightarrow \lambda^2-16 \lambda+55=0\)
\(\Rightarrow(\lambda-5)(\lambda-11)=0\)
\(\Rightarrow \lambda=5,11\)
Hence,
The sum of all possible value of \(\lambda\),
\(\lambda=5+11\)
\(\therefore \quad \lambda =16\)

JEE Main-24.06.2022

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Question Bank Series

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