Answer
Verified
436.5k+ views
Hint: Here, we will rewrite the given equation using the cross product rule, \[a \times \left( {b \times c} \right) = \left( {a \cdot c} \right)b - \left( {a \cdot b} \right)c\] 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 \[\left( {\vec d + \vec a} \right)\left[ {\vec a \times \left( {\vec b \times \left( {\vec c \times \vec d} \right)} \right)} \right]\].
Rewriting the given equation using the cross product rule, \[a \times \left( {b \times c} \right) = \left( {a \cdot c} \right)b - \left( {a \cdot b} \right)c\], we get
\[ \Rightarrow \left( {\vec d + \vec a} \right)\left[ {\vec a \times \left( {\left( {\vec b \cdot \vec d} \right)\vec c - \left( {\vec b \cdot \vec c} \right)\vec d} \right)} \right]\]
Simplifying the above equation by open the open brackets, we get
\[
\Rightarrow \left( {\vec d + \vec a} \right)\left[ {\left( {\vec b \cdot \vec d} \right)\left( {\vec a \times \vec c} \right) - \left( {\vec b \cdot \vec c} \right)\left( {\vec a \times \vec d} \right)} \right] \\
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c + \vec a \cdot \vec a \cdot \vec c} \right] - \left( {\vec b \cdot \vec c} \right)\left[ {\vec a \cdot \vec d \cdot \vec d + \vec a \cdot \vec a \cdot \vec d} \right] \\
\]
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
\[
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c + 0 \cdot \vec c} \right] - \left( {\vec b \cdot \vec c} \right)\left[ {\vec a \cdot 0 + 0 \cdot \vec d} \right] \\
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c + 0} \right] - \left( {\vec b \cdot \vec c} \right)\left[ {0 + 0} \right] \\
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c} \right] - \left( {\vec b \cdot \vec c} \right)\left( 0 \right) \\
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c} \right] - 0 \\
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c} \right] \\
\]
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.
Complete step-by-step answer:
We are given that \[\left( {\vec d + \vec a} \right)\left[ {\vec a \times \left( {\vec b \times \left( {\vec c \times \vec d} \right)} \right)} \right]\].
Rewriting the given equation using the cross product rule, \[a \times \left( {b \times c} \right) = \left( {a \cdot c} \right)b - \left( {a \cdot b} \right)c\], we get
\[ \Rightarrow \left( {\vec d + \vec a} \right)\left[ {\vec a \times \left( {\left( {\vec b \cdot \vec d} \right)\vec c - \left( {\vec b \cdot \vec c} \right)\vec d} \right)} \right]\]
Simplifying the above equation by open the open brackets, we get
\[
\Rightarrow \left( {\vec d + \vec a} \right)\left[ {\left( {\vec b \cdot \vec d} \right)\left( {\vec a \times \vec c} \right) - \left( {\vec b \cdot \vec c} \right)\left( {\vec a \times \vec d} \right)} \right] \\
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c + \vec a \cdot \vec a \cdot \vec c} \right] - \left( {\vec b \cdot \vec c} \right)\left[ {\vec a \cdot \vec d \cdot \vec d + \vec a \cdot \vec a \cdot \vec d} \right] \\
\]
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
\[
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c + 0 \cdot \vec c} \right] - \left( {\vec b \cdot \vec c} \right)\left[ {\vec a \cdot 0 + 0 \cdot \vec d} \right] \\
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c + 0} \right] - \left( {\vec b \cdot \vec c} \right)\left[ {0 + 0} \right] \\
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c} \right] - \left( {\vec b \cdot \vec c} \right)\left( 0 \right) \\
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c} \right] - 0 \\
\Rightarrow \left( {\vec b \cdot \vec d} \right)\left[ {\vec d \cdot \vec a \cdot \vec c} \right] \\
\]
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE