Question

Given $a=i+2j+3k,b=-i+2j+k,$ and $c=3i+j,$ find a vector $r$ which is normal to both $a$ and $b$. What is the inclination of $r$ and $c$?

• A. $\cos \theta =\frac{-4}{\sqrt{30}}.$
• B. $\sin \theta =\frac{-4}{\sqrt{30}}.$
• C. $\sin \theta =\frac{4}{\sqrt{30}}.$
• D. $\cos \theta =\frac{4}{\sqrt{30}}.$

Answer 1

The correct option is A $\cos \theta =\frac{-4}{\sqrt{30}}.$

Resultant $=a+b+c=3i+5j+4k.$

Unit vector in the direction of resultant
$=\frac{3i+5j+4k}{\sqrt{(9+25+16)}}=\frac{1}{5\sqrt{\left(2\right)}}(3i+5j+4k).$
Also $a\times b=-4(i+j+k),$
$\therefore |a\times b|=4\sqrt{3}.$
$(a\times b).c=-4(i+j+k).(3i+j)=-4(3+1)=-16$
$\therefore 4\sqrt{3}.\sqrt{10}\cos \theta =-16$ $\Rightarrow \cos \theta =\frac{-4}{\sqrt{30}}.$