JEE Main 2021 — Mathematics Vectors & 3D Geometry

If two vectors ${a}_{1}\hat{i}+{b}_{1}\hat{j}+{c}_{1}\hat{k}$a1​i^+b1​j^​+c1​k^ and ${a}_{2}\hat{i}+{b}_{2}\hat{j}+{c}_{2}\hat{k}$a2​i^+b2​j^​+c2​k^ are perpendicular, then ${a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0.$a1​a2​+b1​b2​+c1​c2​=0.

Given $\vec{OP}\perp \vec{OQ}$OP⊥OQ​

$$\Rightarrow -x+2y-3x=0$$⇒−x+2y−3x=0

$$\Rightarrow y=2x...(i)$$⇒y=2x...(i)

Also, $\vec{PQ}=\vec{OQ}-\vec{OP}$PQ​=OQ​−OP

$$\Rightarrow \vec{PQ}=(-1-x)\hat{i}+(2-y)\hat{j}+(3x+1)\hat{k}$$⇒PQ​=(−1−x)i^+(2−y)j^​+(3x+1)k^

And $|\vec{PQ}|=\sqrt{20}$∣PQ​∣=20​

$$\Rightarrow \sqrt{{(-1-x)}^{2}+{(2-y)}^{2}+{(3x+1)}^{2}}=\sqrt{20}$$⇒(−1−x)2+(2−y)2+(3x+1)2​=20​

$$\Rightarrow (x+1{)}^{2}+(y-2{)}^{2}+(1+3x{)}^{2}=20$$⇒(x+1)2+(y−2)2+(1+3x)2=20

Put the value from equation $(i),$(i), to get

$$(x+1{)}^{2}+(2x-2{)}^{2}+(1+3x{)}^{2}=20$$(x+1)2+(2x−2)2+(1+3x)2=20

$$\Rightarrow {x}^{2}+2x+1+4{x}^{2}-8x+4+1+6x+9{x}^{2}=20$$⇒x2+2x+1+4x2−8x+4+1+6x+9x2=20

$$\Rightarrow 14{x}^{2}+6=20$$⇒14x2+6=20

$$\Rightarrow 14{x}^{2}=14$$⇒14x2=14

$\Rightarrow x=\pm 1,$⇒x=±1, but given $x>0$x>0

$\Rightarrow x=1$⇒x=1 and $y=2.$y=2.

Now $\vec{OP},\vec{OQ},\vec{OR}$OP,OQ​,OR are coplanar and we know that the three vectors ${a}_{1}\hat{i}+{b}_{1}\hat{j}+{c}_{1}\hat{k},{a}_{2}\hat{i}+{b}_{2}\hat{j}+{c}_{2}\hat{k}$a1​i^+b1​j^​+c1​k^,a2​i^+b2​j^​+c2​k^ and ${a}_{3}\hat{i}+{b}_{3}\hat{j}+{c}_{3}\hat{k}$a3​i^+b3​j^​+c3​k^ are coplanar, then $|\begin{matrix}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{matrix}|=0$∣a1​a2​a3​​b1​b2​b3​​c1​c2​c3​​∣=0

$$\Rightarrow |\begin{matrix}x & y & -1 \\ -1 & 2 & 3x \\ 3 & z & -7\end{matrix}|=0$$⇒∣x−13​y2z​−13x−7​∣=0

$$\Rightarrow |\begin{matrix}1 & 2 & -1 \\ -1 & 2 & 3 \\ 3 & z & -7\end{matrix}|=0$$⇒∣1−13​22z​−13−7​∣=0

$$\Rightarrow 1(-14-3z)-2(7-9)-1(-z-6)=0$$⇒1(−14−3z)−2(7−9)−1(−z−6)=0

$$\Rightarrow z=-2$$⇒z=−2

$$\therefore {x}^{2}+{y}^{2}+{z}^{2}=1+4+4=9.$$∴x2+y2+z2=1+4+4=9.