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If a,b\vec{a}, \vec{b} and c\vec{c} are three unit vectors such that a+2b3c=0\vec{a} + 2\vec{b} - 3\vec{c} = \vec{0}, then the value of 2a.b6b.c3c.a2\vec{a}.\vec{b} - 6\vec{b}.\vec{c} - 3\vec{c}.\vec{a} is

Solution

Correct Option: 1

Since a,b,c\vec{a}, \vec{b}, \vec{c} are unit vectors:

a2=b2=c2=1|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1

The given equation a+2b3c=0\vec{a} + 2\vec{b} - 3\vec{c} = \vec{0} can be rearranged in three ways and squared to find each dot product.


a+2b=3c\vec{a} + 2\vec{b} = 3\vec{c}

Taking the dot product of each side with itself:

a2+4(ab)+4b2=9c2|\vec{a}|^2 + 4(\vec{a}\cdot\vec{b}) + 4|\vec{b}|^2 = 9|\vec{c}|^2

1+4(ab)+4=91 + 4(\vec{a}\cdot\vec{b}) + 4 = 9

ab=1(i)\vec{a}\cdot\vec{b} = 1 \quad \cdots (i)


a=3c2b\vec{a} = 3\vec{c} - 2\vec{b}

Taking the dot product of each side with itself:

a2=9c212(bc)+4b2|\vec{a}|^2 = 9|\vec{c}|^2 - 12(\vec{b}\cdot\vec{c}) + 4|\vec{b}|^2

1=912(bc)+41 = 9 - 12(\vec{b}\cdot\vec{c}) + 4

12(bc)=1212(\vec{b}\cdot\vec{c}) = 12

bc=1(ii)\vec{b}\cdot\vec{c} = 1 \quad \cdots (ii)


2b=3ca2\vec{b} = 3\vec{c} - \vec{a}

Taking the dot product of each side with itself:

4b2=9c26(ca)+a24|\vec{b}|^2 = 9|\vec{c}|^2 - 6(\vec{c}\cdot\vec{a}) + |\vec{a}|^2

4=96(ca)+14 = 9 - 6(\vec{c}\cdot\vec{a}) + 1

6(ca)=66(\vec{c}\cdot\vec{a}) = 6

ca=1(iii)\vec{c}\cdot\vec{a} = 1 \quad \cdots (iii)


Substituting (i)(i), (ii)(ii), and (iii)(iii):

2ab6bc3ca2\vec{a}\cdot\vec{b} - 6\vec{b}\cdot\vec{c} - 3\vec{c}\cdot\vec{a}

=2(1)6(1)3(1)= 2(1) - 6(1) - 3(1)

=263= 2 - 6 - 3

=7= \boxed{-7}

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