Mathematics - Vector Question with Solution | TestHub

Given,

A line passes through $A\left(4,-6,-2\right)$ and $B\left(16,-2,4\right)$,

So, the equation of line will be $\frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6}$

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

So, the direction cosines will be $\frac{6}{\sqrt{6^2+2^2+3^2}},\frac{{\displaystyle 2}}{{\displaystyle \sqrt{6^2+2^2+3^2}}},\frac{{\displaystyle 3}}{{\displaystyle \sqrt{6^2+2^2+3^2}}}$

$\equiv \frac{6}{7},\frac{{\displaystyle 2}}{{\displaystyle 7}},\frac{{\displaystyle 3}}{{\displaystyle 7}}$

Now, given point $P\left(a,b,c\right)$ lies on the line $AB$ and at a distance of $21$ from point $A$

So, by using the equation of line $AB$ we get,

$\Rightarrow \frac{a-4}{{\displaystyle \frac{6}{7}}}=\frac{b+6}{{\displaystyle \frac{2}{7}}}=\frac{c+2}{{\displaystyle \frac{3}{7}}}=21$

$\Rightarrow \left(21\times \frac{6}{7}+4,\frac{2}{7}\times 21-6,\frac{3}{7}\times 21-2\right)\equiv \left(a,b,c\right)$

$\Rightarrow \left(22,0,7\right)\equiv \left(a,b,c\right)$

Hence, the distance between $P\left(22,0,7\right) \& Q\left(4,-12,3\right)$ is given by, $\sqrt{324+144+16}=22$