Vector Algebra Question 385
Question: The value of b such that scalar product of the vectors $ (\mathbf{i}+\mathbf{j}+\mathbf{k}) $ with the unit vector parallel to the sum of the vectors $ (2\mathbf{i}+4\mathbf{j}-5\mathbf{k}) $ and $ (b\mathbf{i}+2\mathbf{j}+3\mathbf{k}) $ is 1, is
[MNR 1992; Roorkee 1985, 95; Kurukshetra CEE 1998; UPSEAT 2000]
Options:
A) ? 2
B) ? 1
C) 0
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
- Parallel vector $ =(2+b)\mathbf{i}+6\mathbf{j}-2\mathbf{k} $ Unit vector $ =\frac{(2+b)\mathbf{i}+6\mathbf{j}-2\mathbf{k}}{\sqrt{b^{2}+4b+44}} $ According to the condition, $ 1=\frac{(2+b)+6-2}{\sqrt{b^{2}+4b+44}} $
$ \Rightarrow b^{2}+4b+44=b^{2}+12b+36 $
$ \Rightarrow 8b=8\Rightarrow b=1. $
BETA
