Question

A vector $a$ is collinear with vector $b=(6,-8,-\frac{15}{2})$ and make an acute angle with the positive direction of $z$-axis. If $|a|=50,$ then $a=$

• A. $(24,32,30)$
• B. $(24,-32,30)$
• C. $(-24,32,30)$
• D. None of these

Answer 1

The correct option is C $(-24,32,30)$

Given $a$ is collinear with $b$. then there exists a scalar $t$ such that $a=tb$

We have $b=6i-8j-\frac{15}{2}k$

thus $a=tb=6ti-8tj-\frac{15t}{2}k$

Now if $a$ makes an angle $\theta$ with the positive $z$-axis then

$\cos \theta =\frac{a.k}{\left|a\right|\left|k\right|}=\frac{\frac{-15t}{2}}{\left|a\right|}$

Since $\theta <{90}^0$ then $\cos \theta =\frac{-15t}{2\left|a\right|}>0\Rightarrow t<0$

Now $\left|a\right|=\sqrt{\{36+64+t^2\left(\frac{225}{4}\right)\}}=30$

$t^2=16\Rightarrow t=-4$ as $t<0$

Hence $a=-24i+32j+30k$

$a:(-24,32,30)$