Vectors 3 Question 18
18. The scalar $\overrightarrow{\mathbf{A}} \cdot[(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}) \times(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}})]$ equals
(1981, 2M)
(a) 0
(b) $[\overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{B}} \overrightarrow{\mathbf{C}}]+\left[\begin{array}{lll}\mathbf{B} & \overrightarrow{\mathbf{C}} & \overrightarrow{\mathbf{A}}\end{array}\right]$
(c) $[\overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{B}} \overrightarrow{\mathbf{C}}]$
(d) None of the above
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Answer:
Correct Answer: 18. (a)
Solution:
- $\overrightarrow{\mathbf{A}} \cdot{(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}) \times(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}})}$
$[\because$ it is a scalar triple product of three vectors of the form $\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}, \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}]$
$$ \begin{aligned} &= \overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{B}} \\ &\quad+\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}}+\overrightarrow{\mathbf{C}} \times \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{C}} \times \overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}} \times \overrightarrow{\mathbf{C}}) \\ &=\overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}})+\overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{C}})+\overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{C}} \times \overrightarrow{\mathbf{B}}) \\ &= {[\overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{B}} \overrightarrow{\mathbf{A}}]-[\overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{B}} \overrightarrow{\mathbf{C}}]=0 } \end{aligned} $$
