MCQ
Let a vector $\vec{\text{r}}$ make angles $60^\circ , 30^\circ$ with it and $y-$axes respectively. Find the angle $\vec{\text{r}}$ make with $z-$axis:
  • A
    $30^\circ$
  • B
    $60^\circ$
  • $90^\circ$
  • D
    $ 120^\circ$

Answer

Correct option: C.
$90^\circ$

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

$X$ and $Y$ are independent events such that $P(X \cap \bar{Y})=\frac{2}{5}$ and $P(X)=\frac{3}{5}$. Then $P(Y)$ is equal to:
Let three vectors $\overrightarrow{ a }, \overrightarrow{ b }$ and $\overrightarrow{ c }$ be such that $\overrightarrow{ c }$ is coplanar with $\overrightarrow{ a }$ and $\overrightarrow{ b }, \overrightarrow{ a } \cdot \overrightarrow{ c }=7$ and $\overrightarrow{ b }$ is perpendicular to $\overrightarrow{ c },$ where $\overrightarrow{ a }=-\hat{ i }+\hat{ j }+\hat{ k }$ and $\overrightarrow{ b }=2 \hat{ i }+\hat{ k },$ then the value of $2|\overrightarrow{ a }+\overrightarrow{ b }+\overrightarrow{ c }|^{2}$ is .........
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$ is:
$\cot\Big(\frac{\pi}{4}-2\cot^{-1}3\Big)=$
The principal value of $\tan^{-1}\Big(\tan\frac{3\pi}{5}\Big)$ is:
Let $ABCD$ be a quadrilateral. If $E$ and $F$ are the mid points of the diagonals $AC$ and $BD$ respectively and $(\overrightarrow{ AB }-\overline{ BC })+(\overrightarrow{ AD }-\overrightarrow{ DC })= k \overline{ FE }$, then $k$ is equal to
If $ a$  and  $ b$  are adjacent sides of a rhombus, then
Let $f(x)=\left\{\begin{array}{l}x-1, x \text { is even, } \\ 2 x, x \text { is odd, }\end{array}\right.$. If for some $a \in N, f(f(f(a)))=21$, then $\lim _{x \rightarrow a^{-}}\left\{\frac{|x|^3}{a}-\left[\frac{x}{a}\right]\right\}$, where $[t]$ denotes the greatest integer less than or equal to $t$, is equal to :
If $A$ and $ B$ are two square matrices such that $B = - {A^{ - 1}}BA$, then ${(A + B)^2} = $
$\int_{}^{} {x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} } \;dx = $