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

Hint: 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.
\[\begin{align}
  & \Rightarrow AB:\dfrac{x-\left( 1-3\mu \right)}{3-\left( 1-3\mu \right)}=\dfrac{y-\left( \mu -1 \right)}{2-\left( \mu -1 \right)}=\dfrac{z-\left( 2+5\mu \right)}{6-\left( 2+5\mu \right)} \\
 & \Rightarrow AB:\dfrac{x-\left( 1-3\mu \right)}{3-1+3\mu }=\dfrac{y-\left( \mu -1 \right)}{2-\mu +1}=\dfrac{z-\left( 2+5\mu \right)}{6-2-5\mu } \\
 & \Rightarrow AB:\dfrac{x-\left( 1-3\mu \right)}{2+3\mu }=\dfrac{y-\left( \mu -1 \right)}{3-\mu }=\dfrac{z-\left( 2+5\mu \right)}{4-5\mu } \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow \left( 2+3\mu \right)\left( 1 \right)+\left( 3-\mu \right)\left( -4 \right)+\left( 4-5\mu \right)\left( 3 \right)=0 \\
 & \Rightarrow 2+3\mu -12+4\mu +12-15\mu =0 \\
 & \Rightarrow -8\mu +2=0 \\
 & \Rightarrow 8\mu =2 \\
\end{align}\]
Now by cross multiplication, then we get
\[\begin{align}
  & \Rightarrow \mu =\dfrac{2}{8} \\
 & \Rightarrow \mu =\dfrac{1}{4}.....(1) \\
\end{align}\]

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 \[\dfrac{d}{a}=\dfrac{b}{e}=\dfrac{c}{f}\].