Let P be the image of the point (3,1,7) with respect to the plane x-y+z=3. Then the equation of the plane passing through P and containing the straight line x \over 1 = y \over 2 = z \over 1 is

3D Geometry PYQ — JEE Advanced 2016 · Shift 2 · Q22 with Solution

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Correct answer: C

Reflect the point $A(3,1,7)$ across the plane

The plane is $x-y+z=3 \quad \Rightarrow \quad x-y+z-3=0.$ So here, $a=1,\; b=-1,\; c=1,\; d=-3.$

If point $A(x_1,y_1,z_1)$ is reflected in the plane $ax+by+cz+d=0$ , then its image $P$ is $P=A-2\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}(a,b,c).$

For $A=(3,1,7)$ , $ax_1+by_1+cz_1+d=3-1+7-3=6,$ and $a^2+b^2+c^2=1^2+(-1)^2+1^2=3.$

Hence, $P=(3,1,7)-2\cdot \frac{6}{3}(1,-1,1)$ $=(3,1,7)-4(1,-1,1)$ $=(-1,5,3).$

So the reflected point is $P=(-1,5,3).$

Interpret the given line

The line is $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}.$ Let the common value be $t$ . Then $x=t,\quad y=2t,\quad z=t.$ So the line passes through the origin and has direction vector $\vec d=(1,2,1).$

Find the plane containing this line and the point $P$

A plane containing the line must contain:

the direction vector of the line: $\vec d=(1,2,1)$ ,
and the vector from a point on the line to $P$ .

Since the line passes through the origin, take $\vec{OP}=(-1,5,3).$

Therefore, a normal vector to the required plane is $\vec n=\vec d\times \vec{OP}.$

Compute:

\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 2 & 1\\ -1 & 5 & 3 \end{vmatrix}$$ $$=\hat i(2\cdot 3-1\cdot 5)-\hat j(1\cdot 3-1\cdot(-1))+\hat k(1\cdot 5-2\cdot(-1))$$ $$=\hat i(6-5)-\hat j(3+1)+\hat k(5+2)$$ $$=(1,-4,7).$$ So the plane has equation $$x-4y+7z=0,$$ since it passes through the origin. --- 4. **Check with options** The required plane is $$x-4y+7z=0,$$ which matches **Option C**. --- 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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View written solutionFree

Correct answer: C

Reflect the point $A(3,1,7)$ across the plane

The plane is $x-y+z=3 \quad \Rightarrow \quad x-y+z-3=0.$ So here, $a=1,\; b=-1,\; c=1,\; d=-3.$

If point $A(x_1,y_1,z_1)$ is reflected in the plane $ax+by+cz+d=0$ , then its image $P$ is $P=A-2\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}(a,b,c).$

For $A=(3,1,7)$ , $ax_1+by_1+cz_1+d=3-1+7-3=6,$ and $a^2+b^2+c^2=1^2+(-1)^2+1^2=3.$

Hence, $P=(3,1,7)-2\cdot \frac{6}{3}(1,-1,1)$ $=(3,1,7)-4(1,-1,1)$ $=(-1,5,3).$

So the reflected point is $P=(-1,5,3).$

Interpret the given line

The line is $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}.$ Let the common value be $t$ . Then $x=t,\quad y=2t,\quad z=t.$ So the line passes through the origin and has direction vector $\vec d=(1,2,1).$

Find the plane containing this line and the point $P$

A plane containing the line must contain:

the direction vector of the line: $\vec d=(1,2,1)$ ,
and the vector from a point on the line to $P$ .

Since the line passes through the origin, take $\vec{OP}=(-1,5,3).$

Therefore, a normal vector to the required plane is $\vec n=\vec d\times \vec{OP}.$

Compute:

\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 2 & 1\\ -1 & 5 & 3 \end{vmatrix}$$ $$=\hat i(2\cdot 3-1\cdot 5)-\hat j(1\cdot 3-1\cdot(-1))+\hat k(1\cdot 5-2\cdot(-1))$$ $$=\hat i(6-5)-\hat j(3+1)+\hat k(5+2)$$ $$=(1,-4,7).$$ So the plane has equation $$x-4y+7z=0,$$ since it passes through the origin. --- 4. **Check with options** The required plane is $$x-4y+7z=0,$$ which matches **Option C**. --- 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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