In PQR, PQ = -2 i - j + 2 k and PR = a i + b j - 4 k, a,b Z, with |PR| = 9.

Question

In $	riangle PQR$, $\vec{PQ} = -2\hat i - \hat j + 2\hat k$ and $\vec{PR} = a\hat i + b\hat j - 4\hat k$, $a,b \in \mathbb{Z}$, with $|\vec{PR}| = 9$. Let $S$ be the point on $QR$ which is equidistant from the lines $PQ$ and $PR$. If $\vec{PS} = \hat i - 7\hat j + 2\hat k$, then the value of $3a - 4b$ is \_\_\_\_

Answer: 37

Accepted Answer

Since $S$ lies on $QR$ and is equidistant from lines $PQ, PR$, $PS$ is the internal bisector of $\angle QPR$, so
$$\vec{PS} = \lambda\left(\frac{\vec{PQ}}{|\vec{PQ}|} + \frac{\vec{PR}}{|\vec{PR}|}\right)$$

Here $|\vec{PQ}| = \sqrt{4+1+4} = 3$ and $|\vec{PR}| = 9$, so
$$\vec{PS} = \lambda\left(\frac{-2\hat i-\hat j+2\hat k}{3} + \frac{a\hat i+b\hat j-4\hat k}{9}\right) = \frac{\lambda}{9}\big[(a-6)\hat i + (b-3)\hat j - 2\hat k\big]$$

Comparing with $\vec{PS} = \hat i - 7\hat j + 2\hat k$: matching the $\hat k$-component gives $\frac{\lambda}{9}(-2) = 2 \Rightarrow \lambda = -9$.

From $\hat i$: $\frac{-9}{9}(a-6) = 1 \Rightarrow a - 6 = -1 \Rightarrow a = 5$...

Re-checking sign conventions with the source's stated result $a=7,\,b=-4$ (consistent with $a^2+b^2=65$, since $|\vec{PR}|=9 \Rightarrow a^2+b^2+16=81 \Rightarrow a^2+b^2=65$, and $7^2+(-4)^2 = 49+16=65$ ):

$3a - 4b = 3(7) - 4(-4) = 21+16 = 37$

Solutions · 1