Question
Choose the correct answer from the given four options.
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are unit vectors such that $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0,$ then the value of $\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}$ is:
  1. 1.
  2. 3.
  3. $-\frac{3}{2}.$
  4. None of these. 

Answer

  1. $-\frac{3}{2}.$

Solution:

We have $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$

$\Rightarrow(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}})(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}})=0$

$\Rightarrow\vec{\text{a}}^2+\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{a}}\cdot\vec{\text{c}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{b}}^2+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}+\vec{\text{c}}\cdot\vec{\text{b}}+\vec{\text{c}}^2=0$

$\Rightarrow\vec{\text{a}}^2+\vec{\text{b}}^2+\vec{\text{c}}^2+2(\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}})=0$ $[\because\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{a}},\vec{\text{b}}\cdot\vec{\text{c}}=\vec{\text{c}}\cdot\vec{\text{b}}\text{ and }\vec{\text{c}}\cdot\vec{\text{a}}=\vec{\text{a}}\cdot\vec{\text{c}}]$

$\Rightarrow1+1+1+2(\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}})=0$

$\Rightarrow\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}=-\frac{3}{2}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Let $R = \{ (3,\,3),\;(6,\;6),\;(9,\,9),\;(12,\,12),\;(6,\,12),\;(3,\,9),(3,\,12),\,(3,\,6)\} $ be a relation on the set $A = \{ 3,\,6,\,9,\,12\} $. The relation is
If $f:[1,\; + \infty ) \to [2,\; + \infty )$ is given by $f(x) = x + \frac{1}{x}$ then ${f^{ - 1}}$ equals
The shortest distance between the lines $\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$ is $......$.
A line passes through the points (6, −7, −1) and (2, −3, 1). The direction cosines of the line so directed that the angle made by it with the positive direction of x-axis is acute, is?
  1. $\frac{2}{3},\frac{2}{3},-\frac{1}{3}$
  2. $-\frac{2}{3},\frac{2}{3},\frac{1}{3}$
  3. $\frac{2}{3}-\frac{2}{3},\frac{1}{3}$
  4. $\frac{2}{3},\frac{2}{3},\frac{1}{3}$
Solution of the differential equation $\cos x\;dy = y\left( {\sin x - y} \right)dx,0 < x < \frac{\pi }{2}$ is
The solution of the differential equation $\frac{{dy}}{{dx}} = (a{e^{bx}} + c\cos mx)$ is
If $A = \left( {\begin{array}{*{20}{c}}1&2&0\\0&1&2\\2&0&1\end{array}} \right),$then $adj \,A$
Consider system of equations  $ x + y -az = 1$  ;  $2x + ay + z = 1$   ; $ax + y -z = 2$
The area bounded by the curve $4 y^{2}=x^{2}(4-x)(x-2)$ is equal to ...... .
The corner points of the feasible region determined by the system of linear constraints are (0, 10),(5, 5),(15, 15),(0, 20). Let z = px + qy where p, q > 0. Condition on p and q so that the maximum of z occurs at both the points (15, 15) and (0, 20) is __________:
  1. q = 2p
  2. p = 2p
  3. p = q
  4. q = 3p