Question

Find a vector of magnitude $\sqrt{51}$ which makes equal angles with the vectors $a=\frac{1}{3}(i-2j+2k),b=\frac{1}{5}(-4i-3k)$ and $c=j$

• A. $\pm (-5i+j+5k).$
• B. $\pm (5i-j-5k).$
• C. $\pm (5i+j-5k).$
• D. $\pm (5i-j+5k).$

Answer 1

The correct option is B $\pm (5i-j-5k).$

Let the required vector be
$a=d_{1}i+d_{2}j+d_{3}k$ where
$d{{}_1}^2+d{{}_2}^2+d{{}_3}^2=51$ (given) ...(1)
Now $\alpha \cdot \beta =\left|\alpha \right|\left|\beta \right|\cos \theta$ or $\cos \theta =\frac{(a.b)}{\left|\alpha \right|\left|\beta \right|}$
Each of the given vectors a, b, c is a unit vector
$\therefore \cos \theta =\frac{d\cdot a}{\left|d\right|\left|a\right|}=\frac{d\cdot b}{\left|d\right|\left|b\right|}=\frac{d\cdot c}{\left|d\right|\left|c\right|}$
or $d.a.=d.b=d.c$ as $\left|d\right|=\sqrt{51}$
cancels out since $\left|a\right|=\left|b\right|=\left|c\right|=1.$
$\therefore \frac{1}{3}(d_1-2d_2+2d_3)=\frac{1}{5}(-4d_1+0d_2-3d_3)=d_2$
$\therefore d_1-5d_2+2d_3=0$ from 1st and 3rd.
$4d_1+5d_2+3d_3=0$ from 2nd and 3rd
$\therefore \frac{d_1}{-15-10}=\frac{d_2}{8-3}=\frac{d_3}{5+20}$
or $\frac{d_1}{5}=\frac{d_2}{-1}=\frac{d_3}{-5}=\lambda ,$ say.
Putting for $d_1,d_2$ and $d_{3}in\left(1\right),$ we get
$(25+1+25){\lambda }^2=51$ $\therefore \lambda =\pm 1$
Hence the required vectors are $\pm (5i-j-5k).$