• # question_answer If a of magnitude 50 is collinear with the vector $\mathbf{b}=6\,\mathbf{i}-8\,\mathbf{j}-\frac{15\,\mathbf{k}}{2},$ and makes an acute angle with the positive direction of z-axis, then the vector a is equal to [Pb. CET 2004] A) $24\,\mathbf{i}-32\,\mathbf{j}+30\,\mathbf{k}$ B) $-24\,\mathbf{i}+32\,\mathbf{j}+30\,\mathbf{k}$ C) $16\,\mathbf{i}-16\,\mathbf{j}-15\,\mathbf{k}$ D) $-12\,\mathbf{i}+16\,\mathbf{j}-30\,\mathbf{k}$

Solution :

• Let $\mathbf{a}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$
• $|\mathbf{a}|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=50$,  $\mathbf{b}=6\mathbf{i}-8\mathbf{j}-\frac{15}{2}\mathbf{k}$
• Since $\mathbf{a}$ and $\mathbf{b}$ are collinear, so $\mathbf{a}=k\,\mathbf{b}$ and $\frac{x}{6}=\frac{y}{-8}=\frac{2z}{-15}=k$,  (constant)
• Þ   $2500={{k}^{2}}\left[ \frac{144+256+225}{4} \right]$
• Þ   $k=\pm \sqrt{\frac{2500\times 4}{625}}=\pm 4$
• Since $\mathbf{a}$makes an acute angle with the direction of
• z-axis, Hence, its z-component must be positive. This is possible only when $k=-4$.
• \ $\mathbf{a}=k\,\left[ 6\mathbf{i}-8\mathbf{j}-\frac{15}{2}\mathbf{k} \right]$,  $[\because \mathbf{a}=k\mathbf{b}]$
• Hence, $\mathbf{a}=-24\mathbf{i}+32\mathbf{j}+30\mathbf{k}$.

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