Forces acting on a particle have magnitude 5, 3 and 1 unit and act in the direction of the vectors 6${\bf{\hat i}} + 2{\bf{\hat j}} + 3{\bf{\hat k}}$, $3{\bf{\hat i}} - 2{\bf{\hat j}} + 6{\bf{\hat k}}$ and $2{\bf{\hat i}} - 3{\bf{\hat j}} - 6{\bf{\hat k}}$ respectively. They remain constant while the particle is displaced from the points $A\left( {2, - 1, - 3} \right)$ to $B\left( {5, - 1,1} \right)$. The work done is

JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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Forces acting on a particle have magnitude 5, 3 and 1 unit and act in the direction of the vectors 6

\mathbf{\hat i} + 2\mathbf{\hat j} + 3\mathbf{\hat k}

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

and

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

respectively. They remain constant while the particle is displaced from the points

A\left( {2, - 1, - 3} \right)

B\left( {5, - 1,1} \right)

. The work done is

View answer and explanation

Correct option: B

33 unit

The correct answer is B (33 unit) because it is the result of calculating the work done by the forces during the displacement, using the dot product of the resultant force and the displacement vector.

More Questions from DOT PRODUCT OF VECTORS

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The distance between the planes given by

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

and

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

is A vector which makes equal angles with the vectors

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

, and

{\hat j}

, is If vector

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

and vector

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

, then

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

is equal to The angle between

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

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

and the projection of

{\mathbf{\vec a}}

in the direction of

{\mathbf{\vec b}}

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

then

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

is equal to A velocity

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

is resolved into two components along

O A

and

O B

making angles

30^{\circ}

and

45^{\circ}

respectively with the given velocity. Then the component along

O B

is Find the equation of the perpendicular drown from the origin to the plane

2x + 4y - 5z = 10