Let \veca=\hati-\hatj+2 \hatk and let \vecb be a vector such that \veca \times \vecb=2 \hati-\hatk and \veca \cdot \vecb=3 . Then the projection of \vecb on the vector \veca-\vecb is :

Vector Algebra PYQ — JEE Main 2022 · 25 Jul · Shift 2 · Q34 with Solution

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Correct answer: A

We have $\vec a = \hat i-\hat j+2\hat k=(1,-1,2)$ and $\vec a\times \vec b = 2\hat i-\hat k=(2,0,-1), \qquad \vec a\cdot \vec b=3.$

We need the projection of $\vec b$ on $\vec a-\vec b$ .

The scalar projection of $\vec b$ on $\vec a-\vec b$ is $\text{proj}_{\vec a-\vec b}(\vec b)=\frac{\vec b\cdot(\vec a-\vec b)}{|\vec a-\vec b|}.$

So we need $\vec b\cdot(\vec a-\vec b)$ and $|\vec a-\vec b|$ .

Use $|\vec a\times \vec b|^2 = |\vec a|^2|\vec b|^2-(\vec a\cdot \vec b)^2.$

First, $|\vec a|^2 = 1^2+(-1)^2+2^2=6.$

Also, $|\vec a\times \vec b|^2 = |(2,0,-1)|^2 = 2^2+0^2+(-1)^2=5.$

Given $\vec a\cdot \vec b=3$ , so $5=6|\vec b|^2-3^2=6|\vec b|^2-9.$ Hence, $6|\vec b|^2=14 \implies |\vec b|^2=\frac{7}{3}.$

$\vec b\cdot(\vec a-\vec b)=\vec a\cdot\vec b-|\vec b|^2=3-\frac{7}{3}=\frac{2}{3}.$

Use $|\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b.$

So, $|\vec a-\vec b|^2=6+\frac{7}{3}-2(3)=6+\frac{7}{3}-6=\frac{7}{3}.$ Thus, $|\vec a-\vec b|=\sqrt{\frac{7}{3}}.$

Therefore,

=\frac{\frac{2}{3}}{\sqrt{\frac{7}{3}}}.$$ Simplify: $$\frac{2}{3}\sqrt{\frac{3}{7}} = \frac{2\sqrt{3}}{3\sqrt{7}} = \frac{2}{\sqrt{21}}.$$ Hence the required projection is $$\boxed{\frac{2}{\sqrt{21}}}.$$ --- 6. **Check options** Option A is $$\frac{2}{\sqrt{21}},$$ which matches our result. So the correct answer is **A**.