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MHTCET 20 Apr 2026 Shift 2 Paper

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Question : 138 of 150

Marks: +1, -0

Let

\vec{a}=2 \hat{i}+\hat{k}, \vec{b}=\hat{i}+\hat{j}+\hat{k}

\vec{c}=4 \hat{i}-3 \hat{j}+7 \hat{k}

\vec{r}

is a vector such that

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

\vec{r} \cdot \vec{a}=0

\vec{r} \cdot \vec{c}=

-14
34
-7
20

Solution:

Concept:

Use the condition

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

to express

\vec{r}

in terms of

\vec{b}

and

\vec{c}

, then apply

\vec{r} \cdot \vec{a} = 0

.

Explanation:

Given

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

.

Rearranging,

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

.

Hence

\vec{r} - \vec{c}

is parallel to

\vec{b}

, so

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

.

Substitute

\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}

and

\vec{b} = \hat{i} + \hat{j} + \hat{k}

.

Thus

\vec{r} = (4+\lambda)\hat{i} + (-3+\lambda)\hat{j} + (7+\lambda)\hat{k}

.

Now

\vec{r} \cdot \vec{a} = 0

with

\vec{a} = 2\hat{i} + \hat{k}

gives:

2(4+\lambda) + (7+\lambda) = 0

.

Simplify:

8 + 2\lambda + 7 + \lambda = 0 \Rightarrow 15 + 3\lambda = 0 \Rightarrow \lambda = -5

.

So

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

.

Finally,

\vec{r} \cdot \vec{c} = (-1)(4) + (-8)(-3) + (2)(7) = -4 + 24 + 14 = 34

.

Answer:

\vec{r} \cdot \vec{c} = 34

, hence option B.

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