Vector Algebra Question 151
Question: 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]
Options:
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} $
Show Answer
Answer:
Correct Answer: B
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}[ \frac{144+256+225}{4} ] $
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,[ 6\mathbf{i}-8\mathbf{j}-\frac{15}{2}\mathbf{k} ] $ , $ [\because \mathbf{a}=k\mathbf{b}] $
Hence, $ \mathbf{a}=-24\mathbf{i}+32\mathbf{j}+30\mathbf{k} $ .
BETA
