3D Geometry Ques 1
- The length of the perpendicular drawn from the point $(2,1,4)$ to the plane containing the lines $\mathbf{r}=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+\lambda(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$ and $\mathbf{r}=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+\mu(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}})$ is
(2019 Main, 12 April II)
(a) $3$
(b) $\frac{1}{3}$
(c) $\sqrt{3}$
(d) $\frac{1}{\sqrt{3}}$
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Answer:
Correct Answer: 1.(c)
Solution: (c) Key Idea $\because$ Length of the perpendicular drawn from point $\left(x_p y_p z_1\right)$ to the plane $a x+b y+c z+d=0$ is
$ d_1=\frac{\left|a x_1+b y_1+c z_1+d\right|}{\sqrt{a^2+b^2+c^2}} $
Given line vectors
$ \begin{aligned} & \mathbf{r}=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+\lambda(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \text { and } \quad ……..(i) \\ & \mathbf{r}=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+\mu(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \quad ……..(ii) \end{aligned} $
Now, a vector perpendicular to the given vectors (i) and (ii) is
$ \begin{aligned} \mathbf{n} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{array}\right| \\ & =\hat{\mathbf{i}}(-4+1)-\hat{\mathbf{j}}(-2-1)+\hat{\mathbf{k}}(1+2) \\ & =-3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $
$\therefore$ The equation of plane containing given vectors (i) and (ii) is
$ \begin{array}{rlrl} \Rightarrow & -3(x-1)+3(y-1)+3(z-0) =0 \\ \Rightarrow & -3 x+3 y+3 z=0 \\ \Rightarrow & x-y-z=0 \end{array} $
Now, the length of perpendicular drawn from the point $(2,1,4)$ to the plane $x-y-z=0$, is
$ \begin{aligned} d_1 & =\frac{|2-1-4|}{\sqrt{1+1+1}} \\ & =\frac{3}{\sqrt{3}}=\sqrt{3} \end{aligned} $
BETA
