Let {a}={i}+{j}, {b}=2 {i}-{k}, {c}=3 {i}-{j}+{k}, then vector {{p}} satisfying {{p}} · {a}=0 and {{p}} × {b}={c} × {b} is

Correct Answer is : i^−j^+2k^{i}-{j}+2 {k}i^−j^+2k^

Correct Answer is : i^−j^+2k^{i}-{j}+2 {k}i^−j^+2k^

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MHTCET 23 Apr 2025 Shift 2 Paper

Question : 115 of 150

Marks: +1, -0

Let

\bar{a}=\hat{i}+\hat{j}, \bar{b}=2 \hat{i}-\hat{k}, \bar{c}=3 \hat{i}-\hat{j}+\hat{k}

\overline{\mathrm{p}}

\overline{\mathrm{p}} \cdot \overline{a}=0

\overline{\mathrm{p}} \times \overline{b}=\overline{c} \times \overline{b}

Solution:

Concept:

\bar{p} \times \bar{b} = \bar{c} \times \bar{b}

, then

\bar{p}-\bar{c}

is parallel to

\bar{b}

, so use a scalar parameter.

Explanation:

Given

\bar{p}\cdot\bar{a}=0

and

\bar{p}\times\bar{b}=\bar{c}\times\bar{b}

From

\bar{p}\times\bar{b}=\bar{c}\times\bar{b}

, we get

(\bar{p}-\bar{c})\times\bar{b}=0

Hence

\bar{p}-\bar{c} = \lambda \bar{b}

, so

\bar{p} = \bar{c} + \lambda \bar{b}

Now substitute in

\bar{p}\cdot\bar{a}=0

(\bar{c}+\lambda\bar{b})\cdot\bar{a}=0

Compute

\bar{c}\cdot\bar{a} = (3\hat{i}-\hat{j}+\hat{k})\cdot(\hat{i}+\hat{j}) = 2

Compute

\bar{b}\cdot\bar{a} = (2\hat{i}-\hat{k})\cdot(\hat{i}+\hat{j}) = 2

2 + 2\lambda = 0

, giving

\lambda = -1

Therefore

\bar{p} = \bar{c} - \bar{b} = (3\hat{i}-\hat{j}+\hat{k}) - (2\hat{i}-\hat{k}) = \hat{i}-\hat{j}+2\hat{k}

Answer:

Option D:

\hat{i}-\hat{j}+2\hat{k}

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