Let \overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b and \overrightarrow c = \widehat j - \widehat k be three vectors such that \overrightarrow a \times \overrightarrow b = \overrightarrow c and \overrightarrow a \,.\,\overrightarrow b = 1 . If the length of projection vector of the vector \overrightarrow b on the vector \overrightarrow a \times \overrightarrow c is l, then the value of 3l2 is equal to \underline\hspace2cm .

Vector Algebra PYQ — JEE Main 2021 · 27 Jul · Shift 1 · Q40 with Solution

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Correct answer: 2

Given vectors

$\vec a=\hat i+\hat j+\hat k=(1,1,1), \qquad \vec c=\hat j-\hat k=(0,1,-1)$

Also, $\vec a\times \vec b=\vec c, \qquad \vec a\cdot \vec b=1.$
Let

$\vec b=(x,y,z).$

Then
$\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 1\\ x & y & z \end{vmatrix} =\hat i(z-y)-\hat j(z-x)+\hat k(y-x).$$ Since this equals $\vec c=(0,1,-1)$, we get \[ z-y=0,\qquad -(z-x)=1,\qquad y-x=-1. \] So, $$z=y, \qquad x-z=1, \qquad y-x=-1.$$ These are consistent.$
Use the dot product condition

$\vec a\cdot \vec b=x+y+z=1.$

Since $z=y$ and $x=y+1$ , $x+y+z=(y+1)+y+y=1$ $3y+1=1$ $y=0.$

Hence, $z=0,\qquad x=1.$

Therefore, $\vec b=(1,0,0)=\hat i.$
Find $\vec a\times \vec c$
$\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 1\\ 0 & 1 & -1 \end{vmatrix} =\hat i(-1-1)-\hat j(-1-0)+\hat k(1-0).$$ So, $$\vec a\times \vec c=-2\hat i+\hat j+\hat k=(-2,1,1).$$$
Projection length of $\vec b$ on $\vec a\times \vec c$

If $\vec d=\vec a\times \vec c$ , then the length of projection of $\vec b$ on $\vec d$ is $l=\frac{|\vec b\cdot \vec d|}{|\vec d|}.$

Now, $\vec b\cdot \vec d=(1,0,0)\cdot(-2,1,1)=-2,$ so $|\vec b\cdot \vec d|=2.$

Also, $|\vec d|=\sqrt{(-2)^2+1^2+1^2}=\sqrt{6}.$

Therefore, $l=\frac{2}{\sqrt6}.$
Compute $3l^2$

$l^2=\frac{4}{6}=\frac{2}{3}$ $3l^2=3\cdot \frac{2}{3}=2.$
Final answer

$\boxed{2}$

View written solutionFree

Correct answer: 2

Given vectors

$\vec a=\hat i+\hat j+\hat k=(1,1,1), \qquad \vec c=\hat j-\hat k=(0,1,-1)$

Also, $\vec a\times \vec b=\vec c, \qquad \vec a\cdot \vec b=1.$
Let

$\vec b=(x,y,z).$

Then
$\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 1\\ x & y & z \end{vmatrix} =\hat i(z-y)-\hat j(z-x)+\hat k(y-x).$$ Since this equals $\vec c=(0,1,-1)$, we get \[ z-y=0,\qquad -(z-x)=1,\qquad y-x=-1. \] So, $$z=y, \qquad x-z=1, \qquad y-x=-1.$$ These are consistent.$
Use the dot product condition

$\vec a\cdot \vec b=x+y+z=1.$

Since $z=y$ and $x=y+1$ , $x+y+z=(y+1)+y+y=1$ $3y+1=1$ $y=0.$

Hence, $z=0,\qquad x=1.$

Therefore, $\vec b=(1,0,0)=\hat i.$
Find $\vec a\times \vec c$
$\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 1\\ 0 & 1 & -1 \end{vmatrix} =\hat i(-1-1)-\hat j(-1-0)+\hat k(1-0).$$ So, $$\vec a\times \vec c=-2\hat i+\hat j+\hat k=(-2,1,1).$$$
Projection length of $\vec b$ on $\vec a\times \vec c$

If $\vec d=\vec a\times \vec c$ , then the length of projection of $\vec b$ on $\vec d$ is $l=\frac{|\vec b\cdot \vec d|}{|\vec d|}.$

Now, $\vec b\cdot \vec d=(1,0,0)\cdot(-2,1,1)=-2,$ so $|\vec b\cdot \vec d|=2.$

Also, $|\vec d|=\sqrt{(-2)^2+1^2+1^2}=\sqrt{6}.$

Therefore, $l=\frac{2}{\sqrt6}.$
Compute $3l^2$

$l^2=\frac{4}{6}=\frac{2}{3}$ $3l^2=3\cdot \frac{2}{3}=2.$
Final answer

$\boxed{2}$