Solveeit Logo

Question

Question: The triple product \[\left( {\vec d + \vec a} \right)\left[ {\vec a \times \left( {\vec b \times \le...

The triple product (d+a)[a×(b×(c×d))]\left( {\vec d + \vec a} \right)\left[ {\vec a \times \left( {\vec b \times \left( {\vec c \times \vec d} \right)} \right)} \right] simplifies to
A.(bd)[dac]\left( {\vec b\vec d} \right)\left[ {\vec d\vec a\vec c} \right]
B.(bc)[abd]\left( {\vec b\vec c} \right)\left[ {\vec a\vec b\vec d} \right]
C.(ba)[abd]\left( {\vec b\vec a} \right)\left[ {\vec a\vec b\vec d} \right]
D.None of these

Explanation

Solution

Here, we will rewrite the given equation using the cross product rule, a×(b×c)=(ac)b(ab)ca \times \left( {b \times c} \right) = \left( {a \cdot c} \right)b - \left( {a \cdot b} \right)c and then simplify it. Then we will use that if in a scalar triple product, two same variables are there, then the value of products is 0 to find the required value.

Complete step-by-step answer:
We are given that (d+a)[a×(b×(c×d))]\left( {\vec d + \vec a} \right)\left[ {\vec a \times \left( {\vec b \times \left( {\vec c \times \vec d} \right)} \right)} \right].

Rewriting the given equation using the cross product rule, a×(b×c)=(ac)b(ab)ca \times \left( {b \times c} \right) = \left( {a \cdot c} \right)b - \left( {a \cdot b} \right)c, we get
(d+a)[a×((bd)c(bc)d)]\Rightarrow \left( {\vec d + \vec a} \right)\left[ {\vec a \times \left( {\left( {\vec b \cdot \vec d} \right)\vec c - \left( {\vec b \cdot \vec c} \right)\vec d} \right)} \right]

Simplifying the above equation by open the open brackets, we get

(d+a)[(bd)(a×c)(bc)(a×d)] (bd)[dac+aac](bc)[add+aad]  \Rightarrow \left( {\vec d + \vec a} \right)\left[ {\left( {\vec b \cdot \vec d} \right)\left( {\vec a \times \vec c} \right) - \left( {\vec b \cdot \vec c} \right)\left( {\vec a \times \vec d} \right)} \right] \\\ \Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c + \vec a \cdot \vec a \cdot \vec c} \right] - \left( {\vec b \cdot \vec c} \right)\left[ {\vec a \cdot \vec d \cdot \vec d + \vec a \cdot \vec a \cdot \vec d} \right] \\\

We know that if in a scalar triple product, two same variables are there, then the value of products is 0.

So, the above expression becomes

(bd)[dac+0c](bc)[a0+0d] (bd)[dac+0](bc)[0+0] (bd)[dac](bc)(0) (bd)[dac]0 (bd)[dac]  \Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c + 0 \cdot \vec c} \right] - \left( {\vec b \cdot \vec c} \right)\left[ {\vec a \cdot 0 + 0 \cdot \vec d} \right] \\\ \Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c + 0} \right] - \left( {\vec b \cdot \vec c} \right)\left[ {0 + 0} \right] \\\ \Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c} \right] - \left( {\vec b \cdot \vec c} \right)\left( 0 \right) \\\ \Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c} \right] - 0 \\\ \Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c} \right] \\\

Hence, option A is correct.

Note: We know that a dot product is the product between components in parallel and cross product is the product between components in perpendicular. These are because of the orthogonal direction in a product only one of the two components (parallel & perpendicular) takes part. The knowledge of both the products of vectors is really important in this question.