Download App

Vector Algebra — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSVector Algebra2 Marks
Question
Find the angle between two vectors $\vec a $ and $\vec b$ with magnitudes 1 and 2 respectively and when $\vec{a} \cdot \vec{b}=1$.
✓

Answer

Given $\vec{a} \cdot \vec{b}=1,|\vec{a}|=1$ and $|\vec{b}|=2$. We have
$\theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a} \| \vec{b}|}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$

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 "2 Marks Questions" from Vector Algebra→Vector Algebra — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Express $\tan ^{-1}( \frac{\cos x}{1-\sin x}),-\frac{3 \pi}{2}<x<\frac{\pi}{2}$ in the simplest form.
→
Evaluate the following integrals:$\int\text{x}^{\text{n}}.\log\text{x dx}$
→
Prove that: $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\frac{1}{2},1} \right]$
→
Can a vector have direction angles 45º, 60º, 120º?
→
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
One of them is black and other is red.
→
Matrix $\text{A}=\begin{bmatrix}0&2\text{b}&-2\\3&1&3\\3\text{a}&3&-1 \end{bmatrix}$ is given to be symmetric, find values of a and b.
→
If $\begin{bmatrix}\text{x}&1 \end{bmatrix}\begin{bmatrix}1&0\\-2&0 \end{bmatrix}=0,$ find x.
→
Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal's selection is $\frac{1}{3}$ and that of Monika's selection is $\frac{1}{5}$. Find the probability that.
Both of them will be selected.
→
Find the second order derivatives of the function given in Exercise:
$\text{x}^3\log\text{x}$
→
Write the value of $\hat{\text{i}}\times\big(\hat{\text{j}}+\hat{\text{k}}\big)+\hat{\text{j}}\times\big(\hat{\text{k}}+\hat{\text{i}}\big)+\hat{\text{k}}\times\big(\hat{\text{i}}+\hat{\text{j}}\big).$
→