A) 8
B) 6
C) 4
D) 2
Correct Answer: A
Solution :
[a] \[\vec{a}+t\vec{b}=(2-t)\hat{i}+(2+2t)\hat{j}+(3+t)\hat{k}\] \[(\vec{a}+t\vec{b})\] and \[\vec{c}\] is perpendicular. Therefore, \[(\vec{a}+t\vec{b}).\vec{c}=0\] \[3(2-t)+2+2t=0\] \[6-3t+2t+2=0\] \[t=8\]You need to login to perform this action.
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