Question

Let $\overrightarrow a ,\;\overrightarrow b ,\;\overrightarrow c$ be three unit vectors such that $\overrightarrow a + 5\overrightarrow b + 3\overrightarrow c = 0$ then $\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)$
is equal to
A. $\overrightarrow a .\overrightarrow b$
B. $\overrightarrow a \left( {.\overrightarrow b + 2\overrightarrow c } \right)$
C. $\overrightarrow b .\left( {\overrightarrow a + \overrightarrow c } \right)$
D.Zero

Answer 1

Hint : To find the required vector, use the given information that is all three vectors are unit vectors and also the equality given. With use of the given equality, put the value of one vector in terms of the other two and then solve the given vector operation to get the required answer.

Complete step by step solution:
In order to find the correct option, we will use the given information to solve the given vector operation,
We have given $\overrightarrow a + 5\overrightarrow b + 3\overrightarrow c = 0$ , we can further write it as
$\Rightarrow \overrightarrow a + 5\overrightarrow b + 3\overrightarrow c = 0 \\\ \Rightarrow \overrightarrow a = - \left( {5\overrightarrow b + 3\overrightarrow c } \right) \;$
Now, putting $\overrightarrow a = - \left( {5\overrightarrow b + 3\overrightarrow c } \right)$ in the given vector operation to solve it
We have to find $\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)$
$\overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right) = - \left( {5\overrightarrow b + 3\overrightarrow c } \right).\left( {\overrightarrow b \times \overrightarrow c } \right)$
Now using the distributive property of dot product, we will get
$- \left( {5\overrightarrow b + 3\overrightarrow c } \right).\left( {\overrightarrow b \times \overrightarrow c } \right) = - \left( {5\overrightarrow b .\left( {\overrightarrow b \times \overrightarrow c } \right) + 3\overrightarrow c .\left( {\overrightarrow b \times \overrightarrow c } \right)} \right)$
Here we can see that there is dot product between $\overrightarrow b \;{\text{and}}\;\left( {\overrightarrow b \times \overrightarrow c } \right),\;{\text{and}}\;\overrightarrow c \;{\text{and}}\;\left( {\overrightarrow b \times \overrightarrow c } \right)$ and we know that cross product of any two vectors give normal vector to their plane as the resultant vector, that is angle between them and the resultant vector will be a right angle. And also we know that the dot product between two perpendicular vectors gives zero. So using these information to solve further, we will get
$\- \left( {5\overrightarrow b .\left( {\overrightarrow b \times \overrightarrow c } \right) + 3\overrightarrow c .\left( {\overrightarrow b \times \overrightarrow c } \right)} \right) = - (5 \times 0 + 3 \times 0) \\\ = 0 \;$
Therefore option D. is correct.
So, the correct answer is “Option D”.

Note : The given set of vectors in this question is linearly independent because the vectors can be expressed as the combination of one another. And the scalar triple product of linearly independent vectors is equal to zero as in this question. Linearly independent shows that all the vectors which are linearly independent with each other lie on the same plane.