Find the angles which the vector a = i - j +2 k makes with the coordinate axes.

Let _ 1 be the angle between a and x-axis. | a |= (1)^2+(-1)^2+(2)^2 =4=2 b = i (Because i is the unit vector along x-axis) | b |=(1)^2=1=1 a . b =1+0+0=1 _ 1 = a . b |a | |b | =1 (2)(1) =1 2 _ 1 = ^ -1 (1 2 )= 3 Let _ 2 be the angle between a and y-axis. | a |= (1)^2+(-1)^2+(2)^2 =4=2 b = j (Because j is the unit vector along y-axis) | b |=(1)^2=1=1 a . b =0-1+0=-1 _ 2 = a . b |a | |b | =-1 (2)(1) =-1 2 _ 2 = ^ -1 (-1 2 )=2 3 Let _ 3 be the angle between a and z-axis. | a |= (1)^2+(-1)^2+(2)^2 =4=2 b = k (Because k is the unit vector along z-axis) | b |=(1)^2=1=1 a . b =0+0+2=2 = a . b |a | |b | = 2 (2)(1) =1 2 = ^ -1 (1 2 )= 4

Find the angles which the vector a = i - j +2 k makes with the co…

Download App

Question

Find the angles which the vector $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\sqrt{2}\hat{\text{k}}$ makes with the coordinate axes.

✓

Answer

Let $\theta_{1}$ be the angle between $\vec{\text{a}}$ and x-axis.
$\big|\vec{\text{a}}\big|=\sqrt{(1)^2+(-1)^2+(\sqrt{2})^2}=\sqrt{4}=2$
$\vec{\text{b}}=\hat{\text{i}}$ (Because $\hat{\text{i}}$ is the unit vector along x-axis)
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2}=\sqrt{1}=1$
$\vec{\text{a}}.\vec{\text{b}}=1+0+0=1$
$\cos\theta_{1}=\frac{\vec{\text{a}} .\vec{\text{b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{1}{(2)(1)}=\frac{1}{2}$
$\Rightarrow\theta_{1}=\cos^{-1}\big(\frac{1}{2}\big)=\frac{\pi}{3}$
Let $\theta_{2}$ be the angle between $\vec{\text{a}}$ and y-axis.
$\big|\vec{\text{a}}\big|=\sqrt{(1)^2+(-1)^2+(\sqrt{2})^2}=\sqrt{4}=2$
$\vec{\text{b}}=\hat{\text{j}}$(Because $\hat{\text{j}}$ is the unit vector along y-axis)
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2}=\sqrt{1}=1$
$\vec{\text{a}}.\vec{\text{b}}=0-1+0=-1$
$\cos\theta_{2}=\frac{\vec{\text{a}} .\vec{\text{b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-1}{(2)(1)}=\frac{-1}{2}$
$\Rightarrow\theta_{2}=\cos^{-1}\big(\frac{-1}{2}\big)=\frac{2\pi}{3}$
Let $\theta_{3}$ be the angle between $\vec{\text{a}}$ and z-axis.
$\big|\vec{\text{a}}\big|=\sqrt{(1)^2+(-1)^2+(\sqrt{2})^2}=\sqrt{4}=2$
$\vec{\text{b}}=\hat{\text{k}}$ (Because $\hat{\text{k}}$ is the unit vector along z-axis)
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2}=\sqrt{1}=1$
$\vec{\text{a}}.\vec{\text{b}}=0+0+\sqrt{2}=\sqrt{2}$
$\cos\theta=\frac{\vec{\text{a}} .\vec{\text{b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{\sqrt{2}}{(2)(1)}=\frac{1}{\sqrt2}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{1}{\sqrt2}\big)=\frac{\pi}{4}$

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 "4 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

If $\text{A} = \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} $ If A =, find A^–1 and hence solve the system of equations x – 2y = 10, 2x + y + 3z = 8 and – 2y + z = 7.

→

Find the shortest distance between the following pairs of lines whose vector equation are:
$\vec{\text{r}}=(\lambda-1)\hat{\text{i}}+(\lambda+1)\hat{\text{j}}-(1+\lambda)\hat{\text{k}}$ and $\vec{\text{r}}=(1-\mu)\hat{\text{i}}+(2\mu-1)\hat{\text{j}}+(\mu+2)\hat{\text{k}}$

→

Write the values of 'a' for which the following distribution of probabilities becomes a probability distributioin:

$\text{X}=\text{x}_\text{i}:$	$-2$	$-1$	$0$	$1$
$\text{P}(\text{X}=\text{x}_\text{i}):$	$\frac{1-\text{a}}{4}$	$\frac{1+2\text{a}}{4}$	$\frac{1-2\text{a}}{4}$	$\frac{1+\text{a}}{4}$