The vector 1 3 ,(2i - 2j + k) is - Vidyadip

MCQ

The vector $\frac{1}{3}\,(2i - 2j + k)$ is

A
A unit vector
B
Perpendicular to the vector $3i + 2j - 2k$
C
Parallel to the vector $ - i + j - \frac{1}{2}k$
✓
All of these

✓

Answer

Correct option: D.

All of these

d
(d) Check it with the options.

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 "M.C.Q (1 Marks)" from STD 12 - 10. vector algebra→STD 12 - 10. vector algebra — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If $y = 3{x^5} + 4{x^4} + 2x + 3$, then

If for $AX = B,$ $B = \left[ \begin{array}{l}9\\52\\0\end{array} \right]$ and ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}3&{ - \frac{1}{2}}&{ - \frac{1}{2}}\\{ - 4}&{\,\,\,\frac{3}{4}}&{\,\,\,\,\frac{5}{4}}\\2&{ - \frac{1}{4}}&{ - \frac{3}{4}}\end{array}} \right]$, then $X$ is equal to

If $P(3,\,4,\,5),$ $Q(4,\,6,\,3),$ $R( - 1,\,2,\,4),$ $S(1,\,0,\,5)$ then the projection of $RS$ on $PQ$ is

The area bounded by the curves $y = x (x - 3)^2$ and $y = x$ is (in sq. units) :

$\int_{ - 1}^1 {x{{\tan }^{ - 1}}x\,dx} $ equals

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}-\text{Ky}=0, \text{y}(0)=1$ approaches to zero when $\text{x}\rightarrow\propto$ if,

$\int_{}^{} {\frac{1}{{({x^2} - 1)\sqrt {{x^2} + 1} }}} \;dx = $

The solution of $\frac{{dy}}{{dx}} + y = {e^{ - x}},\,\,y(0) = 0$, is

${d \over {dx}}\sqrt {{{1 + \cos 2x} \over {1 - \cos 2x}}} = $

The shortest distance between lines $\mathrm{L}_1$ and $\mathrm{L}_2$, where $L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and $L_2$ is the line passing through the points $\mathrm{A}(-4,4,3) \cdot \mathrm{B}(-1,6,3)$ and perpendicular to the line $\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$, is