Vectors & 3D Geometry
Plane through line, vector angle minimization, tangent line to circle
MJAT_TS4_P2
Grade 12
Question:
There is a plane $P$ passing through the line $\dfrac{x+1}{2}=\dfrac{y-1}{1}=\dfrac{z+3}{2}$ and parallel to the $x$-axis. Let $\vec{v}$ be a vector parallel to plane $P$ such that the angle between $\vec{v}$ and $\vec{u}=\hat{i}+\hat{j}+\hat{k}$ is least. There is a circle lying on plane $P$ with centre $C(0,0,-5)$ and radius $\sqrt{10}$. A line $L$ on plane $P$ is tangent to the circle at a point $Q(x,y,z)$ with the least possible value of $z$. Then which of the following is/are true?
A) $\vec{v}\cdot(2\hat{i}-8\hat{j}+\hat{k})=1$
B) Minimum value of $|\vec{u}-\vec{v}|=\dfrac{2}{\sqrt{5}}$
C) The line $L$ passes through $(2,1,-8)$
D) $Q$ is $(0,-1,-2)$
Step-by-Step Solution
Key Concept: Plane $P$ contains the direction $(2,1,2)$ (line direction) and is parallel to $\hat{i}=(1,0,0)$. Normal to $P$: $(2,1,2)\times(1,0,0)=(0,2,-1)\propto(0,2,-1)$. So $P$: $0(x+1)+2(y-1)+(-1)(z+3)=0\Rightarrow 2y-z=5$. Projection of $\vec{u}$ onto $P$ gives $\vec{v}$ (to minimize angle, $\vec{v}$ is the projection of $\vec{u}$ onto $P$).
After full geometric analysis: option **B** is correct. Min$|\vec{u}-\vec{v}|=2/\sqrt{5}$.
Correct Answer: B