MCQ
If $\vec{a}$ is a nonzero vector of magnitude 'a' and $\lambda$ a nonzero scalar, then $\lambda\ \vec{a}$ is unit vector if
  • A
    $\lambda=1$
  • B
    $\lambda=-1$
  • C
    $a=\big|\lambda\big|$
  • $a=1/\big|\lambda\big|$

Answer

Correct option: D.
$a=1/\big|\lambda\big|$
Given: $ \vec{a}$ is a non-zero vector of magnitude a $ \Rightarrow\ \ \ |\vec{a}|=1$
Also given $\lambda\neq0\ \text{and}\ \lambda\vec{a}$ is a unit vector.
$\Rightarrow\ \ |\lambda\vec{a}|=1\ \Rightarrow\ \ |\lambda|\big|\vec{a}\big|=1$
$\Rightarrow\ \ \ \ \ \ |\lambda|a=1\ \ \Rightarrow\ \ a=\frac{1}{|\lambda|}$
Therefore, option (D) is correct.

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

The number of values of $\theta \in (0,\pi)$ for which the system of linear equations
$x + 3y + 7z = 0$
$-x + 4y + 7z = 0$
$(sin\,3\theta )x + (cos\,2\theta )y + 2z = 0$ has a non-trivial solution, is
Matrices $\mathrm{A}$ and $\mathrm{B}$ will be inverse of each other only if
If $f(x) = \frac{{\alpha \,x}}{{x + 1}},\;x \ne - 1$. Then, for what value of $\alpha $ is $f(f(x)) = x$
Find the absolute maximum value of $f(x)=4 x-\frac{1}{2} x^2$ in interval $\left[-2, \frac{9}{2}\right]$.
Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem (using simplex), we find that.
Given that for each $a \in(0,1)$

$\lim _{n \rightarrow 0^{+}} \int_n^{1-n} t^{-3}(1-t)^{a-1} d t$

exists. Let this limit be $g(a)$. In addition, it is given that the function $g(a)$ is differentiable on $(0,1)$.

$1.$ The value of $g\left(\frac{1}{2}\right)$ is

$(A)$ $\pi$ $(B)$ $2 \pi$ $(C)$ $\frac{\pi}{2}$ $(D)$ $\frac{\pi}{4}$

$2.$ The value of $g ^{\prime}\left(\frac{1}{2}\right)$ is

$(A)$ $\frac{\pi}{2}$ $(B)$ $\pi$ $(C)$ $-\frac{\pi}{2}$ $(D)$ $0$

Give the answer question $1$ and $2.$

The function $f(x)=\cot ^{-1} x+x$ increases in the interval
If $x = 2\cos t - \cos 2t ,$ $y = 2\sin t - \sin 2t$, then at $t = {\pi \over 4},{{dy} \over {dx}} = $
The solution of the differential equartion $\frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}=0$ is given by :
The three points $\text{ABC}$ have position vectors $(1, x, 3), (3, 4, 7)$ and $(y, -2, -5)$ are collinear then $(x, y):$