Question

Let $\vec{a} = xi + yj + zk$ and $x = 2y$. If $|\vec{a}| = 5\sqrt{2}$ and $\vec{a}$ makes an angle of $135^\circ$ with the z-axis then $\vec{a} =$

• A. $2\sqrt{3}i + \sqrt{3}j - 3k$
• B. $2\sqrt{6}i + \sqrt{6}j - 6k$
• C. $2\sqrt{5}i + \sqrt{5}j - 5k$
• D. $2\sqrt{5}i + \sqrt{5}j + 5k$

Hint:

Use magnitude and direction angle to form equations. Substitute systematically to solve.

Answer 1

The Correct Option is C

We are given $\vec{a} = xi + yj + zk$ and $x = 2y$. So, $\vec{a} = 2y\,i + y\,j + z\,k$.
Given $|\vec{a}| = 5\sqrt{2}$, we have: $\sqrt{(2y)^2 + y^2 + z^2} = 5\sqrt{2}$
$\Rightarrow \sqrt{5y^2 + z^2} = 5\sqrt{2} \Rightarrow 5y^2 + z^2 = 50$
Also, $\cos(135^\circ) = \frac{z}{|\vec{a}|} = -\frac{1}{\sqrt{2}} \Rightarrow z = -5$
Substituting in the equation: $5y^2 + 25 = 50 \Rightarrow y^2 = 5 \Rightarrow y = \sqrt{5}$
Then $x = 2\sqrt{5}$ and $\vec{a} = 2\sqrt{5}i + \sqrt{5}j - 5k$