Q75 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics

Let $\vec{a} = 2\hat{i} + \hat{k}, \vec{b} = \hat{i} + \hat{j} + \hat{k}$, and $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$, Then $\vec{r} \cdot \vec{c} =$

  • A
    $-14$
  • Correct answer
    B
    $34$
  • C
    $-7$
  • D
    $20$

Solution

Given $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$

$(\vec{r} - \vec{c}) \times \vec{b} = 0$

$\vec{r} - \vec{c} = \lambda \vec{b}$

$\vec{r} = \vec{c} + \lambda \vec{b}$

Taking the dot product with $\vec{a}$:

$\vec{r} \cdot \vec{a} = \vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$

Calculating the dot products:

$\vec{c} \cdot \vec{a} = (4)(2) + (-3)(0) + (7)(1) = 15$

$\vec{b} \cdot \vec{a} = (1)(2) + (1)(0) + (1)(1) = 3$

Substituting these values:

$15 + 3\lambda = 0 \Rightarrow \lambda = -5$

Thus, $\vec{r} = \vec{c} - 5\vec{b}$

$\vec{r} = (4\hat{i} - 3\hat{j} + 7\hat{k}) - 5(\hat{i} + \hat{j} + \hat{k}) = -\hat{i} - 8\hat{j} + 2\hat{k}$

Now, calculating $\vec{r} \cdot \vec{c}$:

$\vec{r} \cdot \vec{c} = (-\hat{i} - 8\hat{j} + 2\hat{k}) \cdot (4\hat{i} - 3\hat{j} + 7\hat{k})$

$\vec{r} \cdot \vec{c} = -4 + 24 + 14 = 34$

Answer: $34$

Hence, the correct answer is option: 'B'.

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