Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Three vectors $\vec{a} ,\vec{b} ,and \: \vec{c}$ are such that $\left | \vec{a} \right |=1 ,\left | \vec{b} \right |=2 , \: \left | \vec{c} \right |=4$ and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$

Then the value of $4\; \vec{a}\cdot \vec{b}+3\; \vec{b}\cdot \; \vec{c}+3\; \vec{c}\cdot \; \vec{a}$ is equal to :

Option 1)
27

Option 2)
-68

Option 3)
-26

Option 4)
- 34

Answers (1)

best_answer

As we learnt in

Scalar Product of two vectors (dot product) -

$\vec{a}\vec{b}=\left | a \right |\left | b \right |Cos\theta$

- wherein

$\Theta$ is the angle between the vectors $\vec{a}\: and\:\vec{b}$

$\left | \vec{a} \right |=1,\left | b \right |=2,\left | \vec{c} \right |=4$

$\vec{a}+\vec{b}+\vec{c}=\vec{0}$

$\vec{a}+\vec{b}=-\vec{c}$

Squaring

$\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta_{1} =\left | \vec{c} \right |^{2}$

$1+4+2\times \vec{a}.\vec{b}=16$

$\vec{a}.\vec{b}=\frac{11}{2}$

$\vec{b}+\vec{c}=-\vec{a}$

$\left | \right\vec{b} |^{2}+\left | \right\vec{c} |^{2}+2\left \right\vec{b}.\left \right\vec{c} =\left | \vec{a} \right |^{2}$

$\vec{b}.\vec{c}\:=\frac{1}{2}\:(1-4-16)\:=\:-\frac{19}{2}$

$\vec{a}+\vec{c}=-\vec{b}\\ \left |\vec{a} \right |^{2} +\left |\vec{c} \right |^{2}+2\vec{a}.\vec{c}=\left | \vec{b} \right |^{2}$

$\vec{a}.\vec{c}=\frac{1}{2}(4-16-1)=-\frac{13}{2}\\ 4\vec{a}.\vec{b}+3\vec{b}.\vec{c}+3\vec{c}.\vec{a}$

$=4\times \frac{11}{2}+3\times -\frac{19}{2}+3\times -\frac{13}{2}=22-3\times 16 =-26$

Option 1)

27

This option is incorrect

Option 2)

-68

This option is incorrect

Option 3)

-26

This option is correct

Option 4)

- 34

This option is incorrect

Posted by

prateek

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’

anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

anam-khan

Low JEE Main rank? Explore alternative paths for engineering admission

Latest Question