Download App

VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA1 Mark
Question
Two or more vectors having the same initial point are:
  1. Coinitial vectors
  2. Colinear vectors
  3. Equal vectors
  4. Cannot say
✓

Answer

  1. Coinitial vectors

Solution:

Two or more vectors having same initial points are known as co-initial vectors.

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 VECTOR ALGEBRA→VECTOR ALGEBRA — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The function $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$
  1. is continuous at x = 0
  2. is not continuous at x = 0
  3. is not continuous at x = 0, but can be made continuous at x = 0
  4. none of these.
→
Which of the termis not used in a linear programming problem:
→
The function ${x^5} - 5{x^4} + 5{x^3} - 1$  is
→
If a drunkard person tries to take a step, then it will be a forward or backward step with probabilities $\frac{1}{4},\frac{1}{2}$ respectively, or he will remain in 'as it is' position. If he tries to take a step $5$ times, then probability that he will be one step away from the initial position
→
Area enclosed between the curve ${y^2}(2a - x) = {x^3}$ and line $x = 2a$ above $x -$ axis is
→
If the points $(2,-3),(k,-1)$ and $(0,4)$ are collinear, then find the value of $4 k$.
→
Integrating factor of the differential equation $\frac{{dy}}{{dx}} + y\tan x - \sec x = 0$ is
→
If $f(x) = \left\{ \begin{gathered} \,[x]\, + \,[ - x],\,\,x \ne 2 \hfill \\ \,\,\,\,\,\,\,\,\,\lambda \,\,\,\,\,\,\,\,\,,\,\,x = \,2\,\,\,\, \hfill \\  \end{gathered}  \right.,$ then $f$ is continuous at $x = 2,$ provided $\lambda$ is (where $[.]$ is $G.I.F.$ )
→
The relation R = {(1, 1), (2, 2), (3, 3)} on the set {1, 2, 3} is:
  1. Symmetric only.
  2. Reflexive only.
  3. An equivalence relation.
  4. Transitive only.
→
If in a binomial distribution $\text{n}=4,\text{P(X}=0)=\frac{16}{81},$ then $\text{P(X}=4)$ equals:
  1. $\frac{1}{16}$
  2. $\frac{1}{81}$
  3. $\frac{1}{27}$
  4. $\frac{1}{8}$
→