A) 2
B) 4
C) 6
D) 8
Correct Answer: D
Solution :
We have, \[\overrightarrow{a}=2\,\hat{i}+2\hat{j}+3\hat{k}\] \[\overrightarrow{b}=-\,\hat{i}+2\hat{j}+\hat{k}\] \[\overrightarrow{c}=3\,\hat{i}+\hat{j}\] Now, \[\overrightarrow{a}+t\,\overrightarrow{b}=(2\hat{i}+2\hat{j}+3\hat{k})+(-t\,\hat{i}+2t\hat{j}+t\,\hat{k})\] \[=(2-t)\hat{i}+(2+2t)\hat{j}+(3+t)\hat{k}\] Since, \[\overrightarrow{a}+t\,\overrightarrow{b}\] is perpendicular to \[\overrightarrow{c}\] \[\therefore \] \[(\overrightarrow{a}+t\,\overrightarrow{b})\cdot \overrightarrow{c}=0\] \[\Rightarrow \]\[\{(2-t)\hat{i}+(2+2t)\hat{j}+(3+t)\hat{k}\}\cdot (3\hat{i}+\hat{j})=0\] \[\Rightarrow \] \[3\,(2-t)+2+2t=0\] \[\Rightarrow \] \[6-3t+2+2t=0\] \[\Rightarrow \] \[t=8\]
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