JEE Main 2015 — Mathematics Vectors & 3D Geometry

Let us change the line into symmetric form.

$$x+y+z+1=0=2x-y+z+3$$x+y+z+1=0=2x−y+z+3

Put $z=1$z=1 , so we get $x+y+2=0$x+y+2=0 and $2x-y+4=0$2x−y+4=0

We will get $x=-2$x=−2, $y=0$y=0

$\therefore$∴ The point $(-2,0,1)$(−2,0,1) lies on the line and perpendicular vector will come from $|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 1\end{matrix}|=2\hat{i}+\hat{j}-3\hat{k}$∣i^12​j^​1−1​k^11​∣=2i^+j^​−3k^

So the equation of line would be $\frac{x+2}{2}=\frac{y}{1}=\frac{z-1}{-3}$2x+2​=1y​=−3z−1​

And the other line $\frac{x-1}{\alpha }=\frac{y+1}{-1}=\frac{z}{1}$αx−1​=−1y+1​=1z​

Shortest distance would be $D=\frac{|(\vec{{a}_{2}}-\vec{{a}_{1}})\cdot (\vec{{b}_{1}}\times \vec{{b}_{2}})|}{|(\vec{{b}_{1}}\times \vec{{b}_{2}})|}$D=∣(b1​​×b2​​)∣∣(a2​​−a1​​)⋅(b1​​×b2​​)∣​

When $\vec{{a}_{1}}=(-2\hat{i}+0\hat{j}+1\hat{k})$a1​​=(−2i^+0j^​+1k^)

$$\vec{{a}_{2}}=(\hat{i}-\hat{j}+0\hat{k})$$a2​​=(i^−j^​+0k^)

$$\vec{{b}_{1}}=2\hat{i}+\hat{j}-3\hat{k}$$b1​​=2i^+j^​−3k^

$$\vec{{b}_{2}}=\alpha \hat{i}-\hat{j}+\hat{k}$$b2​​=αi^−j^​+k^

$D=$D= $|\frac{|\begin{matrix}3 & -1 & -1 \\ 2 & 1 & -3 \\ \alpha & -1 & 1\end{matrix}|}{|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ \alpha & -1 & 1\end{matrix}|}|$∣∣i^2α​j^​1−1​k^−31​∣∣32α​−11−1​−1−31​∣​∣

$$= \frac{|3(1-3)+1(2+3\alpha )+1(2+\alpha )|}{|-2\hat{i}-\hat{j}(2+3\alpha )+\hat{k}(-2-\alpha )|}$$=∣−2i^−j^​(2+3α)+k^(−2−α)∣∣3(1−3)+1(2+3α)+1(2+α)∣​

$\Rightarrow$⇒ $|\frac{-6+2+3\alpha +2+\alpha }{\sqrt{4+{(2+3\alpha )}^{2}+{(2+\alpha )}^{2}}}|=\frac{1}{\sqrt{3}}$∣4+(2+3α)2+(2+α)2​−6+2+3α+2+α​∣=3​1​

$$\frac{|4\alpha -2|}{\sqrt{4+4+12\alpha +9{\alpha }^{2}+4+4\alpha +{\alpha }^{2}}}=\frac{1}{\sqrt{3}}$$4+4+12α+9α2+4+4α+α2​∣4α−2∣​=3​1​

$$|\frac{4\alpha -2}{\sqrt{10{\alpha }^{2}+16\alpha +12}}|=\frac{1}{\sqrt{3}}$$∣10α2+16α+12​4α−2​∣=3​1​

$$\Rightarrow 2\alpha (19\alpha -32)=0$$⇒2α(19α−32)=0

$$\therefore \alpha =\frac{32}{19}$$∴α=1932​