JEE Main 2019 — Mathematics Vectors & 3D Geometry

If two vectors are parallel then they are proportional.

So, let $\vec{{\beta }_{1}}=k\vec{\alpha }$β1​​=kα

Given $\vec{\beta }=\vec{{\beta }_{1}}-\vec{{\beta }_{2}}$β​=β1​​−β2​​

$$\Rightarrow \vec{\alpha }\cdot \vec{\beta }=\vec{\alpha }\cdot \vec{{\beta }_{1}}-\vec{\alpha }\cdot \vec{{\beta }_{2}}...(i)$$⇒α⋅β​=α⋅β1​​−α⋅β2​​...(i)

Now, $\vec{\alpha }\cdot \vec{\beta }=(3\hat{i}+\hat{j})\cdot (2\hat{i}-\hat{j}+3\hat{k})$α⋅β​=(3i^+j^​)⋅(2i^−j^​+3k^)

$$=6-1=5$$=6−1=5

And, $\vec{\alpha }$α is perpendicular to $\vec{{\beta }_{1}},$β1​​, hence $\vec{\alpha }\cdot \vec{{\beta }_{1}}=0$α⋅β1​​=0

And, $\vec{\alpha }\cdot \vec{{\beta }_{2}}=\vec{\alpha }\cdot (k\vec{\alpha })=k{(\vec{\alpha })}^{2}=k{|\vec{\alpha }|}^{2}$α⋅β2​​=α⋅(kα)=k(α)2=k∣α∣2

Also, $|\vec{\alpha }|=\sqrt{{3}^{2}+{1}^{2}}=\sqrt{10}$∣α∣=32+12​=10​

Put, all these values in the equation $(i),$(i), to get $5=k\times 10$5=k×10

$$\therefore k=\frac{1}{2}$$∴k=21​

$$\therefore \vec{{\beta }_{1}}=\frac{1}{2}(3\hat{i}+\hat{j})$$∴β1​​=21​(3i^+j^​)

$$=\frac{3}{2}\hat{i}+\frac{1}{2}\hat{j}$$=23​i^+21​j^​

Now $\vec{{\beta }_{2}}=\vec{{\beta }_{1}}-\vec{\beta }=-\frac{1}{2}\hat{i}+\frac{3}{2}\hat{j}-3\hat{k}$β2​​=β1​​−β​=−21​i^+23​j^​−3k^

$$\therefore \vec{{\beta }_{1}}\times \vec{{\beta }_{2}}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & -3\end{matrix}|$$∴β1​​×β2​​=∣i^23​−21​​j^​21​23​​k^0−3​∣

$$=\hat{i}(-\frac{3}{2}-0)-\hat{j}(-\frac{9}{2}-0)+\hat{k}(\frac{9}{4}+\frac{1}{4})$$=i^(−23​−0)−j^​(−29​−0)+k^(49​+41​)

$$=-\frac{3}{2}\hat{i}+\frac{9}{2}\hat{j}+\frac{5}{2}\hat{k}$$=−23​i^+29​j^​+25​k^

$$=\frac{1}{2}(-3\hat{i}+9\hat{j}+5\hat{k}).$$=21​(−3i^+9j^​+5k^).