A) 3
B) 4
C) 5
D) 2
Correct Answer: C
Solution :
[c] : The distance of the point C from the line \[\vec{r}=\vec{a}+s\vec{b}\] is \[\frac{|(\vec{c}-\vec{a})\times \vec{b}|}{|\vec{b}|}\] Here, \[\vec{a}=\hat{i}-2\hat{j}-\hat{k},\vec{c}=4\hat{i}+2\hat{j}+5\hat{k}\]and \[\vec{b}=6\hat{i}+3\hat{j}+2\hat{k}\] \[\therefore \] \[=-10\hat{i}+30\hat{j}-15\hat{k}=5(-2\hat{i}+6\hat{j}-3\hat{k})\] Hence,\[d=\frac{|(\vec{c}-\vec{a})\times \vec{b}|}{|\vec{b}|}=\frac{5\times 7}{7}=5\].You need to login to perform this action.
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