Let a line L pass through the point P(2,3,1) and be parallel to the line x+3 y-2 z-2=0=x-y+2 z . If the distance of L from the point (5,3,8) is \alpha , then 3 \alpha^2 is equal to :

3D Geometry PYQ — JEE Main 2023 · 30 Jan · Shift 2 · Q41 with Solution

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Correct answer: 158

Find the direction vector of the given line

The line is given as the intersection of the two planes: $x+3y-2z-2=0$ $x-y+2z=0$

A line formed by intersection of two planes has direction vector equal to the cross product of their normals.

Normal to first plane: $\vec n_1=(1,3,-2)$
Normal to second plane: $\vec n_2=(1,-1,2)$

So the direction vector is $\vec d=\vec n_1\times \vec n_2$

Compute:

\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 3 & -2 \\ 1 & -1 & 2 \end{vmatrix}$$ $$\vec d=\hat i(3\cdot 2-(-2)(-1)) - \hat j(1\cdot 2-(-2)(1)) + \hat k(1\cdot (-1)-3\cdot 1)$$ $$\vec d=\hat i(6-2)-\hat j(2+2)+\hat k(-1-3)$$ $$\vec d=(4,-4,-4)$$ This is proportional to $$\vec d=(1,-1,-1)$$ --- 2. **Equation of the required line $L$** Since $L$ passes through $P(2,3,1)$ and is parallel to the above line, its direction vector is $$\vec d=(1,-1,-1)$$. So a vector equation of $L$ is $$\vec r=(2,3,1)+t(1,-1,-1)$$ --- 3. **Distance from point $(5,3,8)$ to line $L$** Let $$A=(2,3,1), \quad Q=(5,3,8)$$ Then $$\overrightarrow{AQ}=Q-A=(5-2,3-3,8-1)=(3,0,7)$$ Distance from point to line is $$\alpha=\frac{|\overrightarrow{AQ}\times \vec d|}{|\vec d|}$$ Now compute $$\overrightarrow{AQ}\times \vec d=(3,0,7)\times (1,-1,-1)$$ $$= \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 0 & 7 \\ 1 & -1 & -1 \end{vmatrix}$$ $$=\hat i(0\cdot (-1)-7\cdot (-1)) - \hat j(3\cdot (-1)-7\cdot 1)+\hat k(3\cdot (-1)-0\cdot 1)$$ $$=\hat i(7)-\hat j(-3-7)+\hat k(-3)$$ $$=(7,10,-3)$$ Hence, $$|\overrightarrow{AQ}\times \vec d|=\sqrt{7^2+10^2+(-3)^2}=\sqrt{49+100+9}=\sqrt{158}$$ Also, $$|\vec d|=\sqrt{1^2+(-1)^2+(-1)^2}=\sqrt{3}$$ Therefore, $$\alpha=\frac{\sqrt{158}}{\sqrt{3}}$$ So, $$\alpha^2=\frac{158}{3}$$ Thus, $$3\alpha^2=158$$ --- 4. **Comparison with stored correct answer** Our derived answer is: $$\boxed{158}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: 158

Find the direction vector of the given line

The line is given as the intersection of the two planes: $x+3y-2z-2=0$ $x-y+2z=0$

A line formed by intersection of two planes has direction vector equal to the cross product of their normals.

Normal to first plane: $\vec n_1=(1,3,-2)$
Normal to second plane: $\vec n_2=(1,-1,2)$

So the direction vector is $\vec d=\vec n_1\times \vec n_2$

Compute:

\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 3 & -2 \\ 1 & -1 & 2 \end{vmatrix}$$ $$\vec d=\hat i(3\cdot 2-(-2)(-1)) - \hat j(1\cdot 2-(-2)(1)) + \hat k(1\cdot (-1)-3\cdot 1)$$ $$\vec d=\hat i(6-2)-\hat j(2+2)+\hat k(-1-3)$$ $$\vec d=(4,-4,-4)$$ This is proportional to $$\vec d=(1,-1,-1)$$ --- 2. **Equation of the required line $L$** Since $L$ passes through $P(2,3,1)$ and is parallel to the above line, its direction vector is $$\vec d=(1,-1,-1)$$. So a vector equation of $L$ is $$\vec r=(2,3,1)+t(1,-1,-1)$$ --- 3. **Distance from point $(5,3,8)$ to line $L$** Let $$A=(2,3,1), \quad Q=(5,3,8)$$ Then $$\overrightarrow{AQ}=Q-A=(5-2,3-3,8-1)=(3,0,7)$$ Distance from point to line is $$\alpha=\frac{|\overrightarrow{AQ}\times \vec d|}{|\vec d|}$$ Now compute $$\overrightarrow{AQ}\times \vec d=(3,0,7)\times (1,-1,-1)$$ $$= \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 0 & 7 \\ 1 & -1 & -1 \end{vmatrix}$$ $$=\hat i(0\cdot (-1)-7\cdot (-1)) - \hat j(3\cdot (-1)-7\cdot 1)+\hat k(3\cdot (-1)-0\cdot 1)$$ $$=\hat i(7)-\hat j(-3-7)+\hat k(-3)$$ $$=(7,10,-3)$$ Hence, $$|\overrightarrow{AQ}\times \vec d|=\sqrt{7^2+10^2+(-3)^2}=\sqrt{49+100+9}=\sqrt{158}$$ Also, $$|\vec d|=\sqrt{1^2+(-1)^2+(-1)^2}=\sqrt{3}$$ Therefore, $$\alpha=\frac{\sqrt{158}}{\sqrt{3}}$$ So, $$\alpha^2=\frac{158}{3}$$ Thus, $$3\alpha^2=158$$ --- 4. **Comparison with stored correct answer** Our derived answer is: $$\boxed{158}$$ This matches the stored correct answer.

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