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Question: If we have an expression as \[\alpha = x(a \times b) + y(b \times c) + z(c \times a)\]and \[\left[ {...

If we have an expression as α=x(a×b)+y(b×c)+z(c×a)\alpha = x(a \times b) + y(b \times c) + z(c \times a)and [abc]=18\left[ {abc} \right] = \dfrac{1}{8} then x+y+zx + y + z is equal to ?
1. 8α(a+b+c)8\alpha (a + b + c)
2. α(a+b+c)\alpha (a + b + c)
3. 8(a+b+c)8(a + b + c)
4. None of these

Explanation

Solution

To solve this question we will use the concept of scalar triple product and for this we need to know about it. A scalar triple product of vectors means taking the dot product of one of the vectors with the cross product of the remaining two. It is denoted as [abc]=(a×b).c\left[ {abc} \right] = (a \times b).c.
The following conclusions can be drawn, by looking into the above formula:
∙\bullet The resultant is always a scalar quantity.
∙\bullet Cross product of the vectors is calculated first, followed by the dot product which gives the scalar triple product.
∙\bullet The physical significance of the scalar triple product formula represents the volume of the parallelepiped whose three coterminous edges represent the three vectors a, b and c.

Complete step-by-step solution:
Properties of Scalar Triple Product:
∙\bullet If the vectors are cyclically permuted, then (a×b).c=a.(b×c)(a \times b).c = a.(b \times c)
∙\bullet The product is cyclic in nature, i.e. a.(b×c)=b.(c×a)=c.(b×c)a.(b \times c) = b.(c \times a) = c.(b \times c)
∙\bullet If the triple product of vectors is zero, then it can be inferred that the vectors are coplanar in nature.
We are given α=x(a×b)+y(b×c)+z(c×a)\alpha = x(a \times b) + y(b \times c) + z(c \times a)
Taking dot product with a on both sides we get ,
α.a=x(a×b).a+y(b×c).a+z(c×a).a\alpha .a = x(a \times b).a + y(b \times c).a + z(c \times a).a
Hence we get ,
α.a=0+y(b×c).a+0\alpha .a = 0 + y(b \times c).a + 0
α.a=y[abc]\alpha .a = y\left[ {abc} \right]
Also we are given that [abc]=18\left[ {abc} \right] = \dfrac{1}{8}
Hence we get y=8α.ay = 8\alpha .a
Similarly α.b=z(c×a).b=z[abc]\alpha .b = z(c \times a).b = z\left[ {abc} \right]
Hence we get z=8α.bz = 8\alpha .b
Again α.c=x(a×b).c\alpha .c = x(a \times b).c
Hence x=8α.cx = 8\alpha .c
Therefore we get x+y+z=8α.c+8α.a+8α.bx + y + z = 8\alpha .c + 8\alpha .a + 8\alpha .b
= 8α(a+b+c)8\alpha (a + b + c)
Therefore option (3) is the correct answer.

Note: To solve such a question one must have a strong grip over the concept of vectors , its related properties and formulae. Perform the calculations in the scalar triple product very carefully. Do not mix the concept of cross product and dot product. Remember that the product is cyclic in nature, i.e. a.(b×c)=b.(c×a)=c.(b×c)a.(b \times c) = b.(c \times a) = c.(b \times c)