3D Geometry 3 Question 25
####25. Let $A$ be a point on the line $\mathbf{r}=(1-3 \mu) \hat{\mathbf{i}}+(\mu-1) \hat{\mathbf{j}}+(2+5 \mu) \hat{\mathbf{k}}$ and $B(3,2,6)$ be a point in the space. Then, the value of $\mu$ for which the vector $\mathbf{A B}$ is parallel to the plane $x-4 y+3 z=1$ is
(a) $\frac{1}{4}$
(b) $-\frac{1}{4}$
(c) $\frac{1}{8}$
(d) $\frac{1}{2}$
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Answer:
(a)
Solution:
- Given equation of line is
$\mathbf{r}=(1-3 \mu) \hat{\mathbf{i}}+(\mu-1) \hat{\mathbf{j}}+(2+5 \mu) \hat{\mathbf{k}}$
Clearly, any point on the above line is of the form $(1-3 \mu, \mu-1,2+5 \mu)$
Let $A$ be $(-3 \mu+1, \mu-1,5 \mu+2)$ for some $\mu \in R$. Then, $\mathbf{A B}=(3-(-3 \mu+1)) \hat{\mathbf{i}}+(2-(\mu-1)) \hat{\mathbf{j}}$
$ \begin{aligned} &+(6-(5 \mu+2)) \hat{\mathbf{k}} \quad[\because \mathbf{A B}=\mathbf{O B}-\mathbf{O A}] \\ &=(3 \mu+2) \hat{\mathbf{i}}+(3-\mu) \hat{\mathbf{j}}+(4-5 \mu) \hat{\mathbf{k}} \end{aligned} $
Normal vector $(\mathbf{n})$ of the plane $x-4 y+3 z=1$ is
$ \mathbf{n}=\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} $
$\because \mathrm{AB}$ is parallel to the plane.
$\therefore \mathbf{n}$ is perpendicular to the $\mathbf{A B}$.
$\Rightarrow \mathbf{A B} \cdot \mathbf{n}=0$
$\Rightarrow[(3 \mu+2) \hat{\mathbf{i}}+(3-\mu) \hat{\mathbf{j}}+(4-5 \mu) \hat{\mathbf{k}}] \cdot[\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}]=0$
[From Eqs. (i) and (ii)]
$ \begin{aligned} & \Rightarrow(3 \mu+2)-4(3-\mu)+3(4-5 \mu)=0 \\ & \Rightarrow \quad-8 \mu+2=0 \\ & \Rightarrow \quad \mu=\frac{1}{4} \end{aligned} $
