Question

The vector from the point $-\hat{i}+2\hat{j}+6\hat{k}$ to the straight line through the point $(2,3,-4)$ and parallel to the vector $6\hat{i}+3\hat{j}+4\hat{k}$, is

• A. $\frac{297}{61}i+\frac{118}{61}j-\frac{534}{61}k$
  
• B. $\frac{297}{61}i-\frac{118}{61}j-\frac{534}{61}k$
• C. $\frac{297}{61}i+\frac{118}{61}j-+\frac{534}{61}k$
• D. none of these

Answer 1

The correct option is A $\frac{297}{61}i+\frac{118}{61}j-\frac{534}{61}k$

Equation of the straight line will be

$\frac{x-2}{6}=\frac{y-3}{3}=\frac{z+4}{4}=t$

Therefore

$x=6t+2$
$y=3t+3$
$z=4t-4$

Therefore, let any point P on the line be

$P=(6t+2,3t+3,4t-4)$

And $A=(-1,2,6)$

Therefore

$\overrightarrow{AP}=(6t+3)i+(3t+1)j+(4t-10)k$

Now for perpendicular distance

$\overrightarrow{AP}.(6i+3j+4k)=0$
Or

$\left(\right(6t+3)i+(3t+1)j+(4t-10\left)k\right).(6i+3j+4k)=0$
$6(6t+3)+3(3t+1)+4(4t-10)=0$
$36t+18+9t+3+16t-40=0$
$61t-19=0$
$t=\frac{19}{61}$

Hence

$\overrightarrow{AP}=\frac{297}{61}i+\frac{118}{61}j-\frac{534}{61}k$

Hence

$d=|\overrightarrow{AP}|$