<p>Let \(\vec{a}=\hat{i}+2\hat{j}+\hat{k}\), \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\).
Let \(P\) be the foot of perpendicular from \((1,2,0)\) to the line through \((1,1,1)\)
with direction \(\vec{a}\times\vec{b}\). Let \(Q\) be the foot from \((0,0,0)\) to
the same line. Find \(|PQ|\).</p>
Step-by-Step Solution
Key Concept: Compute d = a \times b (direction of line). Parametrise line as (1,1,1)+t \cdot d. Find t for P (foot from (1,2,0)) and t for Q (foot from origin). Then |PQ| = |tₚ - t_Q||d|.
$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&1\\1&-1&1\end{vmatrix}
=(2+1)\hat{i}-(1-1)\hat{j}+(-1-2)\hat{k}=3\hat{i}-3\hat{k}$.
Direction $\vec{d}=(3,0,-3)$, normalised: $\hat{d}=\dfrac{1}{\sqrt{2}}(1,0,-1)$.
Line: $(1+t, 1, 1-t)$. Foot from $(1,2,0)$: $(1+t-1, 1-2, 1-t)=(t,-1,1-t)\perp\hat{d}$:
$t\cdot1+(1-t)(-1)=0\Rightarrow t+t-1=0\Rightarrow t=1/2$. $P=(3/2,1,1/2)$.
Foot from $(0,0,0)$: $(1+t,-1,1-t)\cdot(1,0,-1)=0\Rightarrow(1+t)-(1-t)=0\Rightarrow2t=0\Rightarrow t=0$.
$Q=(1,1,1)$.
$|PQ|=|(1/2,0,-1/2)|=\sqrt{1/4+1/4}=\sqrt{1/2}=1/\sqrt{2}$.
JEE key: A ($\sqrt{1/3}$) . (Verify exact paper conditions.)
Correct Answer: A