Let \overrightarrow a = \widehat i - 2\widehat j + 3\widehat k , \overrightarrow b = \widehat i + \widehat j + \widehat k and \overrightarrow c be a vector such that \overrightarrow a + ( \overrightarrow b \times \overrightarrow c ) = \overrightarrow 0 and \overrightarrow b \,.\,\overrightarrow c = 5 . Then the value of 3( \overrightarrow c \,.\,\overrightarrow a ) is equal to \underline\hspace2cm .

Vector Algebra PYQ — JEE Main 2022 · 29 Jun · Shift 2 · Q35 with Solution

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Correct answer: BONUS

$\vec a=(1,-2,3), \qquad \vec b=(1,1,1)$

and

$\vec a+(\vec b\times \vec c)=\vec 0.$

So,

$\vec b\times \vec c=-\vec a=(-1,2,-3).$

Also,

$\vec b\cdot \vec c=5.$

We need to find

$3(\vec c\cdot \vec a).$

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

Then

\begin{vmatrix} \hat i & \hat j & \hat k\\ 1&1&1\\ x&y&z \end{vmatrix} =(z-y)\hat i-(z-x)\hat j+(y-x)\hat k.$$ Thus, $$\vec b\times \vec c=(z-y,\,x-z,\,y-x).$$ Since this equals $(-1,2,-3)$, we get: $$z-y=-1 \quad ...(1)$$ $$x-z=2 \quad ...(2)$$ $$y-x=-3 \quad ...(3)$$ Also from $\vec b\cdot \vec c=5$, $$x+y+z=5. \quad ...(4)$$ --- 3. **Solve the system** From (2): $$x=z+2.$$ From (3): $$y=x-3=z+2-3=z-1.$$ Now check (1): $$z-y=z-(z-1)=1,$$ but (1) requires $$z-y=-1.$$ This is a contradiction. --- 4. **Interpretation** The equations are inconsistent, so **no such vector $\vec c$ exists**. There is also a vector identity reason: For any vectors $\vec b,\vec c$, $$\vec b\cdot(\vec b\times \vec c)=0.$$ But from the given relation, $$\vec b\times \vec c=-\vec a.$$ So this would imply $$\vec b\cdot(-\vec a)=0 \Rightarrow \vec a\cdot \vec b=0.$$ Now, $$\vec a\cdot \vec b=(1)(1)+(-2)(1)+(3)(1)=1-2+3=2\neq 0.$$ Hence the condition itself is impossible. Therefore $\vec c$ does not exist, and the asked quantity $3(\vec c\cdot \vec a)$ is **not defined**. --- 5. **Comparison with stored answer** The stored correct answer is **BONUS**, which indicates the question likely has no valid numerical answer due to inconsistency in the data. So I agree with the stored answer.