JEE Main 2012 — Mathematics Vectors & 3D Geometry

Let $x \hat{i}+y \hat{j}+z \hat{k}$xi^+yj^​+zk^ be the required unit vector. Since $\hat{a}$a^ is perpendicular to $(2 \hat{i}-\hat{j}+2 \hat{k})$(2i^−j^​+2k^). $$\therefore \quad 2 x-y+2 z=0$$∴2x−y+2z=0 Since vector $x \hat{i}+y \hat{j}+z \hat{k}$xi^+yj^​+zk^ is coplanar with the vector $\hat{i}+\hat{j}-\hat{k}$i^+j^​−k^ and $2 \hat{i}+2 \hat{j}-\hat{k}$2i^+2j^​−k^. $$\begin{aligned} \therefore \quad x \hat{i} & +y \hat{j}+z \hat{k} \\ & =p(\hat{i}+\hat{j}-\hat{k})+q(2 \hat{i}+2 \hat{j}-\hat{k}), \end{aligned}$$∴xi^​+yj^​+zk^=p(i^+j^​−k^)+q(2i^+2j^​−k^),​ where $p$p and $q$q are some scalars. $$\begin{aligned} & \Rightarrow x \hat{i}+y \hat{j}+z \hat{k} \\ & =(p+2 q) \hat{i}+(p+2 q) \hat{j}-(p+q) \hat{k} \\ & \Rightarrow x=p+2 q, y=p+2 q, z=-p-q \end{aligned}$$​⇒xi^+yj^​+zk^=(p+2q)i^+(p+2q)j^​−(p+q)k^⇒x=p+2q,y=p+2q,z=−p−q​ Now from equation (i), $$\begin{aligned} & 2 p+4 q-p-2 q-2 p-2 q=0 \\ \Rightarrow & -p=0 \Rightarrow p=0 \\ \therefore & x=2 q, y=2 q, z=-q \end{aligned}$$⇒∴​2p+4q−p−2q−2p−2q=0−p=0⇒p=0x=2q,y=2q,z=−q​ Since vector $x \hat{i}+y \hat{j}+z \hat{k}$xi^+yj^​+zk^ is a unit vector, therefore $$\begin{aligned} & |x \hat{i}+y \hat{j}+z \hat{k}|=1 \\ \Rightarrow & \sqrt{x^2+y^2+z^2}=1 \\ \Rightarrow & x^2+y^2+z^2=1 \\ \Rightarrow & 4 q^2+4 q^2+q^2=1 \end{aligned}$$⇒⇒⇒​∣xi^+yj^​+zk^∣=1x2+y2+z2​=1x2+y2+z2=14q2+4q2+q2=1​ $$\Rightarrow 9 q^2=1 \Rightarrow q=\pm \frac{1}{3}$$⇒9q2=1⇒q=±31​ When $q=\frac{1}{3}$q=31​, then $x=\frac{2}{3}, y=\frac{2}{3}$x=32​,y=32​, $$z=-\frac{1}{3}$$z=−31​ When $q=-\frac{1}{3}$q=−31​, then $x=-\frac{2}{3}, y=-\frac{2}{3}$x=−32​,y=−32​, $$z=\frac{1}{3}$$z=31​ Here required unit vector is $\frac{2}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}$32​i^+32​j^​−31​k^ or $-\frac{2}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{1}{3} \hat{k}$−32​i^−32​j^​+31​k^.