Coplanarity of Two Lines MCQ - Practice Questions & Answers

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Coplanarity of Two Lines is considered one of the most asked concept.
14 Questions around this concept.

Solve by difficulty

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then k is equal to

-1

$\frac{2}{9}$

$\frac{9}{2}$

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Coplanarity of Two Lines

We studied in the previous concept that a line $\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$ to lie in the plane ax + by + cz + d = 0 iff $a l+b m+c n=0 \text { and } a x_{1}+b y_{1}+c z_{1}+d=0$

Thus, the general equation of the plane containing a straight line

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} {\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n} \text { is }} \\\\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}{a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0}\\ \\ \text{where,}\;\;\;\;\;\;{a l+b m+c n=0}$

The equation of the plane containing a straight line $\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$ and parallel to the straight line $\frac{x-x_{2}}{l_{1}}=\frac{y-y_{2}}{m_{1}}=\frac{z-z_{2}}{n_{1}}$ is

$\left|\begin{array}{ccc}{x-x_{1}} & {y-y_{1}} & {z-z_{1}} \\ {l} & {m} & {n} \\ {l_{1}} & {m_{1}} & {n_{1}}\end{array}\right|=0$

Hence, the equation of the plane containing two given straight lines

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}\\\\\text{and}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{x-x_{2}}{l_{1}}=\frac{y-y_{2}}{m_{1}}=\frac{z-z_{2}}{n_{1}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\left|\begin{array}{ccc}{x-x_{1}} & {y-y_{1}} & {z-z_{1}} \\ {l} & {m} & {n} \\ {l_{1}} & {m_{1}} & {n_{1}}\end{array}\right|=0\\\\\\\text{or}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left|\begin{array}{ccc}{x-x_{2}} & {y-y_{2}} & {z-z_{2}} \\ {l} & {m} & {n} \\ {l_{1}} & {m_{1}} & {n_{1}}\end{array}\right|=0$

And, the condition of coplanarity of the given straight lines is given by:

$\left|\begin{array}{ccc}{x_{2}-x_{1}} & {y_{2}-y_{1}} & {z_{2}-z_{1}} \\ {l} & {m} & {n} \\ {l_{1}} & {m_{1}} & {n_{1}}\end{array}\right|=0$

In Vector Form:

If the line $\L_{1}: \vec{\mathbf r}= \vec{\mathbf r}_1+\lambda \vec{\mathbf b}_1 \text{ and } \L_{2}: \vec{\mathbf r}= \vec{\mathbf r}_2+\lambda \vec{\mathbf b}_2$ are coplanar then,

$\left [ \vec{\mathbf r}_1\;\; \vec{\mathbf b}_1\;\;\vec{\mathbf b}_2 \right ]=\left [ \vec{\mathbf r}_2\;\; \vec{\mathbf b}_1\;\;\vec{\mathbf b}_2 \right ]$

and the equation of plane containing them is

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\left [ \vec{\mathbf r}\;\; \vec{\mathbf b}_1\;\;\vec{\mathbf b}_2 \right ]=\left [ \vec{\mathbf r}_1\;\; \vec{\mathbf b}_1\;\;\vec{\mathbf b}_2 \right ]\\\\\text{or}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left [ \vec{\mathbf r}\;\; \vec{\mathbf b}_1\;\;\vec{\mathbf b}_2 \right ]=\left [ \vec{\mathbf r}_2\;\; \vec{\mathbf b}_1\;\;\vec{\mathbf b}_2 \right ]$