The points represented by $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar and $(\sin A)\vec{a} + (2\sin 2B)\vec{b} + (3\sin 3C)\vec{c} - 4\vec{d} = \vec{0}$. Then the least value of $\frac{21}{8}(\sin^2 A + \sin^2 2B + \sin^2 3C)$ is:

  • A
    $1$
  • B
    $2$
  • C
    $4$
  • D
    $3$

Explore More

Similar Questions

If $3 \hat{j}$,$4 \hat{k}$ and $3 \hat{j}+4 \hat{k}$ are the position vectors of the vertices $A$,$B$,and $C$ respectively of $\triangle ABC$,then the position vector of the point in which the bisector of $\angle A$ meets $BC$ is

Find $|\vec{a}|$ and $|\vec{b}|,$ if $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|.$

If $|\bar{a} \times \bar{b}|^2+(\bar{a} \cdot \bar{b})^2=144$ and $|\bar{a}|=4$,then $|\bar{b}|=$

Let $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}$,$\vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$,and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$ and $\vec{r} \cdot (\vec{b}-\vec{c})=0$,then $\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2}$ is equal to:

If $|a+b|=|a-b|$,then

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo