Test Index

MHTCET 10 May 2024 Shift 2 Paper

Question : 117 of 150

Marks: +1, -0

\overline{a}=\hat{\mathrm{j}}-\hat{\mathrm{k}}

\overline{c}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}^{\prime}

\overline{b}

\overline{a} \times \overline{b}+\overline{c}=\overline{0}

\overline{a} \cdot \overline{b}=3

Solution:

Concept:

Use the given cross product and dot product conditions to determine the unknown components of

\overline{b}

Explanation:

Let

\overline{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}

Given

\overline{a}=\hat{j}-\hat{k}=(0,1,-1)

and

\overline{c}=\hat{i}-\hat{j}-\hat{k}=(1,-1,-1)

From

\overline{a}\times\overline{b}+\overline{c}=\overline{0}

, we get

\overline{a}\times\overline{b}=-\overline{c}=(-1,1,1)

Now compute the cross product:

\overline{a}\times\overline{b}=(0,1,-1)\times(b_1,b_2,b_3)=(b_3+b_2,-b_1,-b_1)

Equating components gives

b_3+b_2=-1

and

-b_1=1

, so

b_1=-1

Also,

\overline{a}\cdot\overline{b}=b_2-b_3=3

Solve the system

b_2+b_3=-1

and

b_2-b_3=3

Adding the two equations:

2b_2=2\Rightarrow b_2=1

Substituting back,

b_3=-2

Hence

\overline{b}=-\hat{i}+\hat{j}-2\hat{k}

Answer:

Option A:

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

Go to Question:

Prev Question

Next Question