A) 5
B) 4
C) 6
D) 8
Correct Answer: D
Solution :
Given, \[\overrightarrow{A}=2\hat{i}+2\hat{j}+3\hat{k}\] \[\overrightarrow{B}=-\hat{i}+2\hat{j}+\hat{k}\] and \[\overrightarrow{C}=3\hat{i}+\hat{j}\] \[\overrightarrow{A}+t\overrightarrow{B}=(2\hat{i}+2\hat{j}+3\hat{k})+t(-\hat{i}+2\hat{j}+\hat{k})\] \[=(2-t)\hat{i}+(2+2t)\hat{j}+(3+t)\hat{k}\] \[\because \] \[(\overrightarrow{A}+t\overrightarrow{B})\bot \overrightarrow{C}\] \[\therefore \] \[(\overrightarrow{A}+t\overrightarrow{B}).\overrightarrow{C}=0\] \[\Rightarrow \]\[[(2-t)\hat{i}+(2+2t)\hat{j}+(3+t)\hat{k}].[3\hat{i}+\hat{j}]=0\] \[\Rightarrow \] \[(2-t)3+(2+2t).1=0\] \[\Rightarrow \] \[6-3t+2+2t=0\] \[\Rightarrow \] \[-t+8=0\] \[\Rightarrow \] \[t=8\]
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