Let the position vectors of two points P and Q be 3 i- j+2 k and i+2 j-4 k respectively.

Question

Let the position vectors of two points $P$ and $Q$ be $3\hat i-\hat j+2\hat k$ and $\hat i+2\hat j-4\hat k$ respectively. Let $R$ and $S$ be two points such that the direction ratios of lines $PR$ and $QS$ are $(4,-1,2)$ and $(-2,1,-2)$ respectively. Let lines $PR$ and $QS$ intersect at $T$. If the vector $\vec{TA}$ is perpendicular to both $PR$ and $QS$, and the length of vector $\vec{TA}$ is $\sqrt5$ units, then the modulus of the position vector of $A$ is:

Accepted Answer

$PR:\ (3+4\lambda,\ -1-\lambda,\ 2+2\lambda)$; $QS:\ (1-2\mu,\ 2+\mu,\ -4-2\mu)$. Setting equal:
$$2\lambda+\mu=-1,\qquad \lambda+\mu=-3 \implies \lambda=2,\ \mu=-5.$$
$T=(11,-3,6).$
Direction of $TA$: $(4,-1,2)\times(-2,1,-2)=(0,-4,-2)$.

$A=(11,\ -3-4x,\ 6-2x)$; $|TA|=\sqrt5$:
$$(4x)^2+(2x)^2=5 \implies 20x^2=5 \implies x=\pm\frac12.$$
$$x=\frac12:\ A=(11,-5,5);\qquad x=-\frac12:\ A=(11,-1,7).$$
Both give $|A|=\sqrt{121+25+25}=\sqrt{171}$ (and $\sqrt{121+1+49}=\sqrt{171}$).

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