Let a be a vector which is perpendicular to the vector 3 i +1 2 j… - Vidyadip

MCQ

Let $\vec{a}$ be a vector which is perpendicular to the vector $3 \hat{ i }+\frac{1}{2} \hat{ j }+2 \hat{ k }$. If $\overrightarrow{ a } \times(2 \hat{ i }+\hat{ k })=2 \hat{ i }-13 \hat{ j }-4 \hat{ k }$, then the projection of the vector $\vec{a}$ on the vector $2 \hat{ i }+2 \hat{ j }+\hat{ k }$ is

A
$\frac{1}{3}$
B
$1$
✓
$\frac{5}{3}$
D
$\frac{7}{3}$

✓

Answer

Correct option: C.

$\frac{5}{3}$

c
$(\vec{a} \times(2 \hat{i}+\hat{k})) \times\left(3 \hat{i}+\frac{1}{2} \hat{j}+2 \hat{k}\right)$

$=(2 \hat{i}-13 \hat{j}-4 \hat{k}) \times\left(3 \hat{i}+\frac{1}{2} \hat{j}+2 \hat{k}\right)$

$-(6+2) \overrightarrow{ a }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & -13 & -4 \\ 3 & \frac{1}{2} & 2\end{array}\right|$

$\vec{a}=3 \hat{i}+2 \hat{j}-5 \hat{k}$

Projection of $\vec{a}$ on vector $2 \hat{i}+2 \hat{j}+\hat{k}$ is

$\overrightarrow{ a } \cdot \frac{(2 \hat{ i }+2 \hat{ j }+\hat{ k })}{3}=\frac{5}{3}$

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 10. vector algebra→10. 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

Let $[t]$ be the greatest integer less than or equal to $t$. Let $A$ be the set of al prime factors of $2310$ and $f: A \rightarrow \mathbb{Z}$ be the function $f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right]$. The number of one-to-one functions from $A$ to the range of $f$ is :

If $u = 2\,i + 2j - k$ and $v = 6\,i - 3\,j + 2\,k,$ then a unit vector perpendicular to both $u$ and $ v $ is

Let $f: R \rightarrow R$ be a differentiable function such that its derivative $f^{\prime}$ is continuous and $f(\pi)=-6$. If $F:[0, \pi] \rightarrow R$ is defined by $F(x)=\int_0^{ x } f( t ) dt$, and if $\int_0^\pi\left(f^{\prime}( x )+ F ( x )\right) \cos x dx =2$ then the value of $f(0)$ is. . . . . .

Choose the correct answer from the given four options.
If $\tan^{-1}\text{x}+\tan^{-1}\text{y}=\frac{4\pi}{5},$ then $\cot^{-1}\text{x}+\cot^{-1}\text{y}$ equals to:

$\frac{\pi}{5}$
$\frac{2\pi}{5}$
$\frac{3\pi}{5}$
$\pi$

The angle of intersection of the curves xy = a² and x² - y² = 2a²is:

0°
45°
90°
None of these.

The equation of a curve passing through $(1, 0)$ for which the product of the abscissa of a point $P \,\&$ the intercept made by a normal at $P$ on the $x-$ axis equals twice the square of the radius vector of the point $P,$ is

Statement-$1$ : ${\cot ^{ - 1}}\left[ {\frac{{\log \left( {e/{x^2}} \right)}}{{\log \left( {e{x^2}} \right)}}} \right] + {\cot ^{ - 1}}\left[ {\frac{{\log (e{x^2})}}{{\log (e/{x^2})}}} \right]$ = $\frac {\pi}{2}$

Statement-$2$ : ${\tan ^{ - 1}}\left[ {\frac{{1 + \log {x^2}}}{{1 - \log {x^2}}}} \right]$ = ${\tan ^{ - 1}}\,1 + \,{\tan ^{ - 1}}\left( {\log {x^2}} \right)$

Let $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x},$ where $b$ is a constant such that $0 < b < 1$. Then

Let ${R_1}$ be a relation defined by ${R_1} = \{ (a,\,b)|a \ge b,\,a,\,b \in R\} $. Then ${R_1}$ is

$\int_{}^{} {u\frac{{{d^2}v}}{{d{x^2}}}dx - \int_{}^{} {v\frac{{{d^2}u}}{{d{x^2}}}dx = } } $