JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

1 Easy

1 of 1

Application +4 | -1

Two forces

\overrightarrow{{P}}=3 \hat{{i}}-2 \hat{{j}}+\hat{{k}}

and

\overrightarrow{{Q}}=\hat{{i}}+3 \hat{{j}}-5 \hat{{k}}

acting on a particle

A

move it to

B

. If the position vectors of

A

and

B

are respectively

-2 \hat{{i}}+5 \hat{{k}}

and

3 \hat{{i}}-7 \hat{{j}}+2 \hat{{k}}

, then the total work done is

View answer and explanation

Correct option: D

25 unit

Option D (25 unit) is correct because the work done is calculated by finding the dot product of the total force and the displacement vector.

More Questions from DOT PRODUCT OF VECTORS

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\mathbf{r} .(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})+5=0

\mathbf{r}-(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})-8=0

\frac{1}{3}\left( {\hat i - 2\hat j + 2\hat k} \right),\frac{1}{5}\left( { - 4\hat i - 3\hat k} \right)

{\hat j}

\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

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

\frac{\text { Projection of vector } \mathbf{a} \text { on vector } \mathbf{b}}{\text { Projection of vector } \mathbf{b} \text { on vector } \mathbf{a}}

{\mathbf{\vec a}}

{\mathbf{\vec b}}

\frac{{5\pi }}{6}

{\mathbf{\vec a}}

{\mathbf{\vec b}}

\frac{{ - 6}}{{\sqrt 3 }}

\left| {\mathbf{\vec a}} \right|

\frac{1}{4} \mathrm{~m} / \mathrm{s}

O A

O B

30^{\circ}

45^{\circ}

O B

2x + 4y - 5z = 10