JEE Main 2019 — Mathematics Vectors & 3D Geometry

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

We know that the vector joining two points $({x}_{1},{y}_{1},{z}_{1})$(x1​,y1​,z1​) and $({x}_{2},{y}_{2},{z}_{2})$(x2​,y2​,z2​) is $({x}_{2}-{x}_{1})\hat{i}+({y}_{2}-{y}_{1})\hat{j}+({z}_{2}-{z}_{1})\hat{k},$(x2​−x1​)i^+(y2​−y1​)j^​+(z2​−z1​)k^,

Hence, the vector $\vec{AP}=(-1-2)\hat{i}+(2-3)\hat{j}+(6+4)\hat{k}$AP=(−1−2)i^+(2−3)j^​+(6+4)k^

$$\Rightarrow \vec{AP}=-3\hat{i}-\hat{j}+10\hat{k}$$⇒AP=−3i^−j^​+10k^

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

We know that the projection of a vector $\vec{a}$a on a vector $\vec{b}$b is $\frac{|\vec{a}\cdot \vec{b}|}{|\vec{b}|}$∣b∣∣a⋅b∣​

Hence, $AD=\frac{|\vec{AP}\cdot \vec{n}|}{|\vec{n}|}$AD=∣n∣∣AP⋅n∣​

$$\Rightarrow AD=\frac{(-3\hat{i}-\hat{j}+10\hat{k})\cdot (6\hat{i}+3\hat{j}-4\hat{k})}{\sqrt{{6}^{2}+{3}^{2}+{(-4)}^{2}}}$$⇒AD=62+32+(−4)2​(−3i^−j^​+10k^)⋅(6i^+3j^​−4k^)​

$$\Rightarrow AD=\frac{|-18-3-40|}{\sqrt{36+9+16}}$$⇒AD=36+9+16​∣−18−3−40∣​

$\Rightarrow AD=\sqrt{61}$⇒AD=61​ units

The distance between the points $({x}_{1},{y}_{1},{z}_{1})$(x1​,y1​,z1​) and $({x}_{2},{y}_{2},{z}_{2})$(x2​,y2​,z2​) is $\sqrt{{({x}_{1}-{x}_{2})}^{2}+{({y}_{1}-{y}_{2})}^{2}+{({z}_{1}-{z}_{2})}^{2}}$(x1​−x2​)2+(y1​−y2​)2+(z1​−z2​)2​

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

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

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