MHT CET · Maths · Vector Algebra
If the constant forces \(2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\) and \(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) act on a particle due to which it is displaced from a point \(A(4,-3,-2)\) to a point \(B(6,1,-3)\), then the work done by the forces is
- A 15 unit
- B 9 unit
- C \(-15\) unit
- D \(-9\) unit
Answer & Solution
Correct Answer
(C) \(-15\) unit
Step-by-step Solution
Detailed explanation
Resultant force,
\(
\overrightarrow{\mathbf{F}}=(2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})+(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=\hat{\mathbf{i}}-3 \mathbf{j}+5 \hat{\mathbf{k}}
\)
Displacement,
\(\overrightarrow{\mathbf{d}}=\overrightarrow{\mathbf{A B}} =(6 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})-(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \)
\( =2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}} \)
\( W =\overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{d}}=(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \)
\( =2-12-5=-15 \text { unit}\)
\(
\overrightarrow{\mathbf{F}}=(2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})+(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=\hat{\mathbf{i}}-3 \mathbf{j}+5 \hat{\mathbf{k}}
\)
Displacement,
\(\overrightarrow{\mathbf{d}}=\overrightarrow{\mathbf{A B}} =(6 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})-(4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \)
\( =2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}} \)
\( W =\overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{d}}=(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \)
\( =2-12-5=-15 \text { unit}\)
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