The distance of the point having position vector - i ^ + 2 j ^ + 6 k ^ from the straight line passing…

Let, the given point be $P\equiv -\hat{i}+2\hat{j}+6\hat{k}$ and $A\equiv \left(2, 3, -4\right),$ hence the position vector of the point $A$ is $2\hat{i}+3\hat{j}-4\hat{k}.$

We know that the vector joining two points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)$ is $\left(x_2-x_1\right)\hat{i}+\left(y_2-y_1\right)\hat{j}+\left(z_2-z_1\right)\hat{k},$

Hence, the vector $\overrightarrow{AP}=\left(-1-2\right)\hat{i}+\left(2-3\right)\hat{j}+\left(6+4\right)\hat{k}$

$\Rightarrow \overrightarrow{AP}=-3\hat{i}-\hat{j}+10\hat{k}$

And, $AD$ is the projection of $\overrightarrow{AP}$ on the given line i.e. $6\hat{i}+3\hat{j}-4\hat{k}$

We know that the projection of a vector $\overrightarrow{a}$ on a vector $\overrightarrow{b}$ is $\frac{\left|\overrightarrow{a}·\overrightarrow{b}\right|}{\left|\overrightarrow{b}\right|}$

Hence, $AD=\frac{\left|\overrightarrow{AP}\cdot \overrightarrow{n}\right|}{\left|\overrightarrow{n}\right|}$

$\Rightarrow AD=\frac{\left(-3\hat{i}-\hat{j}+10\hat{k}\right)·\left(6\hat{i}+3\hat{j}-4\hat{k}\right)}{\sqrt{6^2+3^2+{\left(-4\right)}^2}}$

$\Rightarrow AD=\frac{\left|-18-3-40\right|}{\sqrt{36+9+16}}$

$\Rightarrow AD=\sqrt{61}$ units

The distance between the points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)$ is $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

Thus, $AP=\sqrt{{\left(-1-2\right)}^2+{\left(2-3\right)}^2+{\left(6+4\right)}^2}=\sqrt{110}$ units

Now, in triangle $ADP,$ by Pythagoras theorem, we have ${\left(AP\right)}^2={\left(PD\right)}^2+{\left(AD\right)}^2$

$\Rightarrow PD=\sqrt{{\left(AP\right)}^2-{\left(AD\right)}^2}=\sqrt{110-61}=7$ units.