Question

Let A be a point on the line $\vec{r}=\left( 1-3\mu \right)\hat{i}+\left( \mu -1 \right)\hat{j}+\left( 2+5\mu \right)\hat{k}$ and $B\left( 3,2,6 \right)$ be a point in the space. Then the value of $\mu$ for which the vector $A\vec{B}$ is parallel to the plane $x-4y+3z=1$?

Answer 1

If $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ be two points on a line AB, then the equation of line AB is $L:\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}$. We know that $L:\dfrac{x-{{x}_{1}}}{d}=\dfrac{y-{{y}_{1}}}{e}=\dfrac{z-{{z}_{1}}}{f}$ is parallel to plane $P:ax+by+cz+k=0$ if $ad+be+cf=0$. By using these concepts, we can find the value of $\mu$ for which the vector $A\vec{B}$ is parallel to the plane $x-4y+3z=1$.

Complete step by step answer:
From the question, we were given that to find the value $\mu$ for which the vector $A\vec{B}$ is parallel to the plane $x-4y+3z=1$.
Let us assume a point $A\left( \left( 1-3\mu \right),\left( \mu -1 \right),\left( 2+5\mu \right) \right)$ on the line $\vec{r}=\left( 1-3\mu \right)\hat{i}+\left( \mu -1 \right)\hat{j}+\left( 2+5\mu \right)\hat{k}$. We were also given a point $B\left( 3,2,6 \right)$.
If $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ be two points on a line AB, then the equation of line AB is $L:\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}$.
So, now we should find a line equation AB which passes through point $A\left( \left( 1-3\mu \right),\left( \mu -1 \right),\left( 2+5\mu \right) \right)$ and point $B\left( 3,2,6 \right)$.
Now we should find the line equation AB.

& \Rightarrow AB:\dfrac{x-\left( 1-3\mu \right)}{3-\left( 1-3\mu \right)}=\dfrac{y-\left( \mu -1 \right)}{2-\left( \mu -1 \right)}=\dfrac{z-\left( 2+5\mu \right)}{6-\left( 2+5\mu \right)} \\\ & \Rightarrow AB:\dfrac{x-\left( 1-3\mu \right)}{3-1+3\mu }=\dfrac{y-\left( \mu -1 \right)}{2-\mu +1}=\dfrac{z-\left( 2+5\mu \right)}{6-2-5\mu } \\\ & \Rightarrow AB:\dfrac{x-\left( 1-3\mu \right)}{2+3\mu }=\dfrac{y-\left( \mu -1 \right)}{3-\mu }=\dfrac{z-\left( 2+5\mu \right)}{4-5\mu } \\\ \end{aligned}$$ We know that $$L:\dfrac{x-{{x}_{1}}}{d}=\dfrac{y-{{y}_{1}}}{e}=\dfrac{z-{{z}_{1}}}{f}$$ is parallel to plane $$P:ax+by+cz+k=0$$ if $$ad+be+cf=0$$. From the question, it was given that the line equation AB is parallel to $$x-4y+3z=1$$. Then we get $$\begin{aligned} & \Rightarrow \left( 2+3\mu \right)\left( 1 \right)+\left( 3-\mu \right)\left( -4 \right)+\left( 4-5\mu \right)\left( 3 \right)=0 \\\ & \Rightarrow 2+3\mu -12+4\mu +12-15\mu =0 \\\ & \Rightarrow -8\mu +2=0 \\\ & \Rightarrow 8\mu =2 \\\ \end{aligned}$$ Now by cross multiplication, then we get $$\begin{aligned} & \Rightarrow \mu =\dfrac{2}{8} \\\ & \Rightarrow \mu =\dfrac{1}{4}.....(1) \\\ \end{aligned}$$ **From equation (2), it is clear that the value of $$\mu $$ for which the vector $$A\vec{B}$$ is parallel to the plane $$x-4y+3z=1$$ is equal to $$\dfrac{1}{4}$$.** **Note:** Students may have a misconception that $$L:\dfrac{x-{{x}_{1}}}{d}=\dfrac{y-{{y}_{1}}}{e}=\dfrac{z-{{z}_{1}}}{f}$$ is parallel to plane $$P:ax+by+cz+k=0$$ if $$\dfrac{d}{a}=\dfrac{b}{e}=\dfrac{c}{f}$$. But we know that $$L:\dfrac{x-{{x}_{1}}}{d}=\dfrac{y-{{y}_{1}}}{e}=\dfrac{z-{{z}_{1}}}{f}$$ is parallel to plane $$P:ax+by+cz+k=0$$ if $$ad+be+cf=0$$. So, students should have a clear view of the concept. Students should also know that $$L:\dfrac{x-{{x}_{1}}}{d}=\dfrac{y-{{y}_{1}}}{e}=\dfrac{z-{{z}_{1}}}{f}$$ is perpendicular to plane $$P:ax+by+cz+k=0$$ if $$\dfrac{d}{a}=\dfrac{b}{e}=\dfrac{c}{f}$$.