Forces of magnitudes 3 and 4 units acting along $6\widehat {\bf{i}} + 2\widehat {\bf{j}} + 3{\bf{\hat k}}$ and $3\widehat {\bf{i}} - 2\widehat {\bf{j}} + 6{\bf{\hat k}}$ respectively act on a particle and displace it from $\left( {2,2 - 1} \right)$ to $\left( {4,3,1} \right)$. The work done is

JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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Application +4 | -1

Forces of magnitudes 3 and 4 units acting along

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

and

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

respectively act on a particle and displace it from

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

\left( {4,3,1} \right)

. The work done is

View answer and explanation

Correct option: A

124/7

Option A (

124/7

) is correct because it is the result of calculating the work done by the forces over the given displacement using the dot product.

Practice All Chapter MCQs →

\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