Question

If a, b, c are the three non-coplanar vectors and p, q, r are defined by the relations \(\vec p = \frac{{\left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}},\vec q = \frac{{\left( {\vec c \times \vec a} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}},\vec r = \frac{{\left( {\vec a \times \vec b} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}}\)then \(\left( {\vec a + \vec b} \right)\;.\;\vec p\; + \left( {\vec b + \vec c} \right)\;.\;\vec q\; + \left( {\vec c + \vec a} \right)\;.\;\vec r = ?\)
A)\[0\]
B) \[1\]
C) \[2\]
D) \[3\]

Answer 1

Hint: In this question we are going to use the algebra of vectors. Use vector multiplication and addition. Solve each part of the equation individually and after that add all these to get required value. Non coplanar vectors are those vectors which are not present in the same plane or parallel planes.If two vectors in a scalar triple product are the same then the scalar triple product will be zero.



Formula Used:\(\left( {\vec a + \vec b} \right).\vec c = \vec a.\vec c + \vec b.\vec c\) {Scalar product is distributive in nature}
\(\left( {\vec a + \vec b} \right){\rm{.}}\vec c = \vec c{\rm{.}}\left( {\vec a + \vec b} \right)\)




Complete step by step solution:Given: Three vectors p, q and r are non coplanar and \(\vec p = \frac{{\left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}},\vec q = \frac{{\left( {\vec c \times \vec a} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}},\vec r = \frac{{\left( {\vec a \times \vec b} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}}\)
\(\left( {\vec a + \vec b} \right).\vec p = \vec a.\vec p + \vec b.\vec p\)
\(\vec a.\vec p + \vec b.\vec p\; = \overrightarrow {a.} \frac{{\left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}} + \;\overrightarrow {b.} \frac{{\left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}}\)
\(\overrightarrow {a.} \frac{{\left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}} + \;\overrightarrow {b.} \frac{{\left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}} = \frac{{\overrightarrow {a.} \left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}} + \frac{{\overrightarrow {b.} \left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}}\)
\(\frac{{\overrightarrow {a.} \left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}} + \frac{{\overrightarrow {b.} \left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}} = \frac{{\left[ {\vec a\vec b\vec c} \right]}}{{\left[ {\vec a\vec b\vec c} \right]}} + \frac{{\left[ {\vec b\vec b\vec c} \right]}}{{\left[ {\vec a\vec b\vec c} \right]}}\)



We know that:
If two vectors in a scalar triple product are the same then the scalar triple product will be zero.
Above statement in mathematical form can be written as
\(\left[ {\vec b\vec c\vec b} \right] = 0\)
\(\frac{{\overrightarrow {a.} \left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}} + \frac{{\overrightarrow {b.} \left( {\vec b \times \vec c} \right)}}{{\left[ {\vec a\vec b\vec c} \right]}} = \frac{{\left[ {\vec a\vec b\vec c} \right]}}{{\left[ {\vec a\vec b\vec c} \right]}} + \frac{{\left[ {\vec b\vec b\vec c} \right]}}{{\left[ {\vec a\vec b\vec c} \right]}} = 1 + 0\)
Similarly, we can find
\(\left( {\vec b + \vec c} \right){\rm{.}}\overrightarrow {q\;} = \left( {\vec c + \vec a} \right){\rm{.}}\overrightarrow {r\;} = 1\)
\(\left( {\vec a + \vec b} \right){\rm{.}}\vec p + \left( {\vec b + \vec c} \right){\rm{.}}\vec q + \left( {\vec c + \vec a} \right){\rm{.}}\vec r = 1 + 1 + 1 = 3\)



Option ‘D’ is correct

Note: Here in this question we have to find the value of the given vector equation. We will use the algebra of vectors i.e. multiplication and addition of vectors in order to find required value.
Vector addition follows commutative and associative law.
Triangle law and parallelogram law are two methods of vector addition.
Vector products are distributive in nature.
Don’t try to solve the whole equation together because it will take more time and there will be a chance of mistakes.