A force of magnitude 5 unit acting along the vector $2{\bf{\hat i}} - 2{\bf{\hat j}} + {\bf{\hat k}}$ displaces the point of applications from (1,2,3) to (5,3,7) then the work done is

JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

1 Easy

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Application +4 | -1

A force of magnitude 5 unit acting along the vector

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

displaces the point of applications from (1,2,3) to (5,3,7) then the work done is

View answer and explanation

Correct option: B

50/3 unit

Option B (50/3 unit) is correct because it is the result of the dot product of the force vector and the displacement vector.

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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