Download App

STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
If $a, b, c$  are mutually perpendicular unit vectors, then $|a + b + c|\,\, = $
  • ✓
    $\sqrt 3 $
  • B
    $3$
  • C
    $1$
  • D
    $0$
✓

Answer

Correct option: A.
$\sqrt 3 $
a
(a) Three mutually perpendicular unit vectors $ = a$, $b$ and $c$.

Therefore $|a|\, = \,|b|\, = \,|c|\, = 1$ and $a.b = b.c = c.a = 0$.

We know that

$|a + b + c{|^2} = (a + b + c)\,.\,(a + b + c) = \,\,|a{|^2} + |b{|^2}$

$ + |c{|^2} + 2(a\,.\,b\,\, + b\,.\,c\, + c\,.\,a) = 1 + 1 + 1 + 0 = 3$

or $|a + b + c|\, = \sqrt 3 .$

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 the matrix $A=\left(\begin{array}{cc}0 & 2 \\ K & -1\end{array}\right)$ satisfies $A\left(A^{3}+3 I\right)=2 I$ then the value of $\mathrm{K}$ is :
→
If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}$ where $a>b>0,$ then $\frac{d x}{d y}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ is
→
Derivative of $e^{\sin ^2 x}$ with respect to $\cos x$ is
→
Evaluate: $\int\frac{1-\cos\text{x}}{cos\text{x}(1+cos\text{x})}\text{dx}.$
→
A line makes angles $\alpha,\beta,\gamma$ with the positive direction of the axes of reference. The value of $\cos^2\alpha+\cos^2\beta+\cos^2\gamma$ is:
→
The function

$f(x)=\frac{4 x^{3}-3 x^{2}}{6}-2 \sin x+(2 x-1) \cos x$

→
$\text{f}(\text{x})=\sin +\sqrt{3}\cos\text{x}$ is maximum when x =
→
The function $f : R \rightarrow R$ defined by $f(x) = (x - 1)(x - 2)(x - 3)$ is:
→
$\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$ equals:
→
The number of distinct real values of $\lambda $ on for which the lines $\frac{{x - 1}}{1} = \frac{{y - 2}}{2} = \frac{{z + 3}}{{{\lambda ^2}}}$ and $\frac{{x - 3}}{1} = \frac{{y - 2}}{{{\lambda ^2}}} = \frac{{z - 1}}{2}$ are coplanar is
→