3D Geometry 3 Question 20

####20. The plane containing the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3}$ and also containing its projection on the plane $2 x+3 y-z=5$, contains which one of the following points?

(2019 Main, 11 Jan I)

(a) $(-2,2,2)$

(b) $(2,2,0)$

(c) $(2,0,-2)$

(d) $(0,-2,2)$

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Answer:

(c)

Solution:

  1. Let the direction vector of the line

$ \frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3} \text { is } \mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \text {. } $

Since, the required plane contains this line and its projection along the plane $2 x+3 y-z=5$, it will also contain the normal of the plane $2 x+3 y-z=5$.

Normal vector of the plane $2 x+3 y-z=5$ is $\mathbf{n}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$.

Now, the required plane contains $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\mathbf{n}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$.

$\therefore$ Normal of the required plane is $\mathbf{b} \times \mathbf{n}$.

Since, the plane contains the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3}$, therefore it also contains the point $\mathbf{a}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$.

Now, the equation of required plane is $(\mathbf{r}-\mathbf{a}) \cdot(\mathbf{b} \times \mathbf{n})=0$

$ \left|\begin{array}{ccc} x-3 & y+2 & z-1 \\ 2 & -1 & 3 \\ 2 & 3 & -1 \end{array}\right|=0 $

$\Rightarrow(x-3)[1-9]-(y+2)[-2-6]+(z-1)[6+2]=0$

$\Rightarrow-8 x+8 y+8 z+32=0$

$\Rightarrow x-y-z=4$

Note that $(2,0,-2)$ is the only point which satisfy above equation.



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