Vector Algebra Question 160
Question: If $ \overrightarrow{A}=\mathbf{i}+2\mathbf{j}+3\mathbf{k},,\overrightarrow{B}=-\mathbf{i}+2\mathbf{j}+\mathbf{k} $ and $ \overrightarrow{C}=3\mathbf{i}+\mathbf{j}, $ then the value of t such that $ \overrightarrow{A}+t\overrightarrow{B} $ is at right angle to vector $ 3\mathbf{i}+4\mathbf{j} $ is
[RPET 2002]
Options:
A) 2
B) 4
C) 5
D) 6
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \overrightarrow{A}+t,\overrightarrow{B}=(\mathbf{i}+2\mathbf{j}+3\mathbf{k})+t(-\mathbf{i}+2\mathbf{j}+\mathbf{k}) $
$ =\mathbf{i}(1-t)+\mathbf{j}(2+2t)+\mathbf{k}(3+t) $
But it is perpendicular to $ \overrightarrow{C}=3\mathbf{i}+\mathbf{j}, $
So, $ \overrightarrow{C}.,(\overrightarrow{A}+t,\overrightarrow{B})=0\Rightarrow 3(1-t)+2+2t=0\Rightarrow t=5. $
BETA
