If |a|=2,|b|=3|c|=4 then [a+b b+c c-a] is equal to - Vidyadip

MCQ

If $|\bar{a}|=2,|\bar{b}|=3|\bar{c}|=4$ then $[\bar{a}+\bar{b} \bar{b}+\bar{c} \bar{c}-\bar{a}]$ is equal to

A
24
B
-24
✓
0
D
48

✓

Answer

Correct option: C.

0

$[\bar{a}+\bar{b} \bar{b}+\bar{c} \bar{c}-\bar{a}]=0$

$Explanation:$
$(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}-\bar{a})] \\
=\bar{b} \times \bar{c}-\bar{b} \times \bar{a}+\bar{c} \times \bar{c}-\bar{c} \times \bar{a} \\
=(\bar{a}+\bar{b}) \cdot[\bar{b} \times \bar{c}-\bar{b} \times \bar{a}-\bar{c} \times \bar{a}] \\
=\bar{a} \cdot(\bar{b} \times \bar{c})-\bar{a} \cdot(\bar{b} \times \bar{a})-\bar{a} \cdot(\bar{c} \times \bar{a})+\bar{b} \cdot(\bar{b} \times \bar{c})-\bar{b} \cdot(\bar{b} \times \bar{a})-\bar{b} \cdot(\bar{c} \times \bar{a}) \\
=\bar{a} \cdot(\bar{b} \times \bar{c})+0+0+0+0-\bar{b} \cdot(\bar{c} \times \bar{a}) \\
=(\bar{a} \bar{b} \bar{c})-(\bar{a} \bar{b} \bar{c})=0$

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

Explore more

More "MCQ" from Vectors→Vectors — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

The sides of a triangle are $4 cm, 5 cm$ and 6 cm . The area of the triangle is equal to

If $\text{f(x)}=\frac{1}{1-\text{x}},$ then the set of points discontinuity of the function $f(f(f(x)))$ is:

$\bar{a}$ and $\bar{b}$ are two non-collinear vectors, then $x \overline{ a }+y \overline{b}$ (where $x$ and $y$ are scalars) represents a vector which is

If $g(x)=x^2+x-2$ and $\frac{1}{2} g \circ f(x)=2 x^2-5 x+2$, then $f(x)$ is equal to:

The area under the curve y = 2√x, enclosed between the lines x = 0 and x = 1 is

A random variable X has the following probability distribution:

X = x	-3	-2	-1	0	1	2	3
P$(X = x)$	0.05	0.10	0.15	0.20	0.25	0.15	0.10

Then $\mathrm{P}(\mathrm{X}$ is odd $)=$

Assertion (A) : If $(-1,3,2)$ and $(5,3,2)$ are respectively the orthocentre and circumcentre of a triangle, then $(3,3,2)$ is its centroid.
Reason (R) : Centroid of a triangle divides the line segment joining the orthocentre and the circumcentre in the ratio $1: 2$.

If $I=\int \frac{d x}{\sin (x-a) \sin (x-b)}$, then $I$ is given by

Let the function $f : R - \{-b\} \rightarrow R - {1}$ be defined by $\text{f(x)}=\frac{\text{x}+\text{a}}{\text{x}+\text{b}},\ \text{a}\neq\text{b}.$ Then,

If $\cos ^{-1}\left(\frac{1}{x}\right)=\theta$, then $\tan \theta=$