If the lines vecr= (hati-hatj+hatk )+lambda (3hatj-hatk ) and vecr= (alphahati-hatj )+mu (2hati-3hatk ) are co-planar, then the distance of the plane containing these two lines from the point (alpha,0,0) is :

3D Geometry: If the lines vecr= (hati-hatj+hatk )+lambda (3hatj-hatk ) and vecr=… | JEE Main 2022

JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1

If the lines $\vec{r}=\left(\hat{i}-\hat{j}+\hat{k}\right)+\lambda\left(3\hat{j}-\hat{k}\right)$ and $\vec{r}=\left(\alpha\hat{i}-\hat{j}\right)+\mu\left(2\hat{i}-3\hat{k}\right)$ are co-planar, then the distance of the plane containing these two lines from the point $(\alpha,0,0)$ is :

3D Geometry

$\frac{2}{9}$$\frac{2}{11}$✓ Correct answer$\frac{4}{11}$$2$

Solution

Points $(1,-1,1)$, $(\alpha,-1,0)$; directions $(0,3,-1)$, $(2,0,-3)$. Coplanar: $\begin{vmatrix}\alpha-1&0&-1\\0&3&-1\\2&0&-3\end{vmatrix}=3(-3(\alpha-1)+2)=0\Rightarrow\alpha=\tfrac53$. Normal $=(0,3,-1)\times(2,0,-3)=(-9,-2,-6)\parallel(9,2,6)$. Plane through $(1,-1,1)$: $9x+2y+6z=13$. Distance from $(\tfrac53,0,0)$: $\frac{|15-13|}{\sqrt{121}}=\frac2{11}$. Option (B).

3D Geometry - JEE Mathematics question (JEE Main 2022 (Online) 26th June Evening Shift)

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