Answer
(a) $d=\frac{|a \times b|}{|a|}$
(b) $\sqrt {\frac{97}{3}}$
Work Step by Step
(a) Area of a parallelogram produced by vectors a and b is equal to the magnitude of the cross product between the vectors.
Base of this parallelogram is $|a|$ and the height is $d$.
$A=|a \times b|$
$A= bh$
$A=|a| d$
$A=|a \times b|$
$|a| d=|a \times b|$
$d=\frac{|a \times b|}{|a|}$
(b) $P(1,1,1), Q(0,6,8), R(-1,4,7)$
$a=QR= \lt -1, -2,-1 \gt$
$b=QP= \lt 1,-5,-7 \gt$
Cross product $a \times b= \lt -1, -2,-1 \gt \times \lt 1,-5,-7 \gt= \lt 9,-8,7 \gt$
From part (a),
$d=\frac{|a \times b|}{|a|}$
$d= \frac{\sqrt{9^2+(-8)^2+7^2 }}{\sqrt{(-1)^2+(-2)^2+(-1)^2 }}$
$=\frac{\sqrt {194}}{\sqrt 6}$
$=\sqrt {\frac{97}{3}}$