3D Geometry 3 Question 24
####24. The plane passing through the point $(4,-1,2)$ and parallel to the lines $\frac{x+2}{3}=\frac{y-2}{-1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3}$ also passes through the point
(a) $(-1,-1,-1)$
(b) $(1,1,-1)$
(c) $(1,1,1)$
(d) $(-1,-1,1)$
(2019 Main, 10 Jan I)
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Answer:
(c)
Solution:
- Let $\mathbf{a}$ be the position vector of the given point $(4,-1,2)$.
Then, $\mathbf{a}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
The direction vector of the lines
$\frac{x+2}{3}=\frac{y-2}{-1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3}$
are respectively
$ \begin{aligned} \mathbf{b} _ {1}=3 \hat{\mathbf{i}} - \hat{\mathbf{j}} + 2 \hat{\mathbf{k}} \\ \text { and } \quad \mathbf{b} _ {2} =\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}} \end{aligned} $
Now, as the plane is parallel to both $\mathbf{b}{1}$ and $\mathbf{b}{2}$
$[\because$ plane is parallel to the given lines]
So, normal vector (n) of the plane is perpendicular to both $\mathbf{b}{1}$ and $\mathbf{b}{2}$.
$\Rightarrow \quad \mathbf{n}=\mathbf{b}_ {1} \times \mathbf{b}_{2}$ and
Required equation of plane is
$ \begin{aligned} & (\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0 \\ & \Rightarrow \quad(\mathbf{r} - \mathbf{a}) \cdot\left(\mathbf{b} _ {1} \times \mathbf{b} _ {2}\right) = 0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} x - 4 & y + 1 & z - 2 \\ 3 & - 1 & 2 \\ 1 & 2 & 3 \end{array}\right|=0 \\ & \because \mathbf{r} - \mathbf{a} = (x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}) - (4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ & = (x - 4) \hat{\mathbf{i}} + (y + 1) \hat{\mathbf{j}}+(z-2) \hat{\mathbf{k}} \end{aligned} $
and we know that, $[\mathbf{a} \mathbf{b} \mathbf{c}]=\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})$
$ =\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right| $
$\Rightarrow(x-4)(-3-4)-(y+1)(9-2)+(z-2)(6+1)=0$
$\Rightarrow \quad-7(x-4)-7(y+1)+7(z-2)=0$
$\Rightarrow \quad(x-4)+(y+1)-(z-2)=0$
$\Rightarrow \quad x+y-z-1=0$
$(1,1,1)$ is the only point that satisfies.
