Question

If b and c are any two no-collinear unit vectors and a is any vector, then
\[\left( a\centerdot b \right)b+\left( a\centerdot c \right)c+\dfrac{\left( a\centerdot \left( b\times c \right) \right)}{{{\left| b\times c \right|}^{2}}}{{\left( b\times c \right)}^{{}}}\] is equal to
(a) 0
(b) a
(c) b
(d) c

Answer 1

Hint: We will be using vector algebra to solve the problem. We will also be using scalar product, dot product and vector triple product to simplify the problem.

Complete step-by-step answer:

We have been given that b and c are two non-collinear unit vectors and a is any vector.

Now, we have to find the value of expression \[\left( a\centerdot b \right)b+\left( a\centerdot c \right)c+\dfrac{\left( a\centerdot \left( b\times c \right) \right)}{{{\left| b\times c \right|}^{2}}}\left( b\times c \right)\].

Now, we know that a unit vector along a vector \[\overset{\to }{\mathop{c}}\,\] can be written as \[\dfrac{\overset{\to }{\mathop{c}}\,}{\left| \overset{\to }{\mathop{c}}\, \right|}\] so we will write \[\dfrac{\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,}{\left| \overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\, \right|}\] as a unit vector.

 \[\left( a\centerdot b \right)b+\left( a\centerdot c \right)c+\dfrac{\left( a\centerdot \left( b\times c \right) \right)}{\left| b\times c \right|}\dfrac{\left( b\times c \right)}{\left| b\times c \right|}\]

\[\left( a\centerdot b \right)b+\left( a\centerdot c \right)c+\left( \dfrac{a\centerdot \left( b\times c \right)}{\left| b\times c \right|} \right)\dfrac{\left( b\times c \right)}{\left| b\times c \right|}\]

Now, we know that a, b are non-collinear unit vectors also \[\left( a\times b \right)\] is perpendicular to both a and b by definition. So, we can say that a, b, \[\dfrac{\left( b\times c \right)}{\left| b\times c \right|}\] are three non-collinear unit vectors. So, we can have now \[\left( a\centerdot b \right)b\] the projection of a along b.

Similarly, \[\left( a\centerdot c \right)c\] is projection of a along c and \[\left( \dfrac{a\centerdot \left( b\times c \right)}{\left| b\times c \right|} \right)\dfrac{\left( b\times c \right)}{\left| b\times c \right|}\] is projection of a along \[\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,\] so the term becomes

\[\left( a-b \right)b+\left( a\centerdot c \right)c+\left( a\centerdot \left( \dfrac{\left( b\times c \right)}{\left| b\times c \right|} \right) \right)\left( b\times c \right)\ =\ {{a}_{b}}\widehat{b}+{{a}_{c}}\widehat{c}+{{a}_{d}}\widehat{d}\]

Where,

\[{{a}_{b}}\ =\ a\centerdot b\]

\[{{a}_{c}}\ =\ a\centerdot c\]

\[{{a}_{d}}\ =\ a\centerdot \left( \dfrac{b\times c}{\left| b\times c \right|} \right)\]

\[\widehat{a}\ =\ a\]

\[\widehat{b}\ =\ b\]

\[\widehat{d}\ =\ \dfrac{b\times c}{\left| b\times c \right|}\]

Now, we know that any z vector can be represented in terms of any three collinear vectors x, y and k as

\[z\ =\ {{z}_{x}}\widehat{x}+{{z}_{y}}\widehat{y}+{{z}_{k}}\widehat{k}\]

Then, we can see that the expression is none other than vector \[\overset{\to }{\mathop{a}}\,\] itself written in terms of b, c, \[\dfrac{b\times c}{\left| b\times c \right|}\] .
So, \[a\ =\ \left( a\centerdot b \right)b+\left( a\centerdot c \right)c+\left( \dfrac{a\centerdot \left( b\times c \right)}{\left| b\times c \right|} \right)\dfrac{\left( b\times c \right)}{\left| b\times c \right|}\]

Hence, the correct option is (b).

Note: To solve these types of questions one must have a good understanding of vectors and vector algebra also. We must note that we have used, fact that a vector can be represented in terms of any three non-collinear vectors.