# The distance of the point having position vector $-\hat{i}+2\hat{j}+6\hat{k}$from the straight line passing through the point ( 2 , 3 , -4 ) and parallel to the vector, $6\hat{i}+3\hat{j}-4\hat{k}$ is : Option 1) 7 Option 2) $4\sqrt3$ Option 3) 6 Option 4) $2\sqrt{13}$

Point P(-1,2,6) & A(2,3,-4)

and $\vec{n}=6\hat{i}+3\hat{j}-4\hat{k}$

$\vec{x}= 6\hat{i}+3\hat{j}-4\hat{k}$

$PD^{2}=AP^{2}-AD^{2}$

$AD=|\frac{\vec{AP}.\vec{n}}{|\vec{n}|}|$

$AD=|\frac{(3\hat{i}+\hat{j}-10\hat{k}).(6\hat{i}+3\hat{j}-4\hat{k})}{|6\hat{i}+3\hat{j}-4\hat{k}|}|$

$=\sqrt{61}$

$PD=\sqrt{110-61}=7$

CORRECT OPTION IS (1).

Option 1)

7

Option 2)

$4\sqrt3$

Option 3)

6

Option 4)

$2\sqrt{13}$

Exams
Articles
Questions