(D) Given: $|\vec{a}|=1, |\vec{b}|=2$ and $|\vec{a}-\vec{b}|^2+|\vec{a}+2\vec{b}|^2=20$. Expanding the dot products: $(\vec{a}-\vec{b}) \cdot (\vec{a}

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(D) Given: $|\vec{a}|=1, |\vec{b}|=2$ and $|\vec{a}-\vec{b}|^2+|\vec{a}+2\vec{b}|^2=20$. Expanding the dot products: $(\vec{a}-\vec{b}) \cdot (\vec{a}-\vec{b}) + (\vec{a}+2\vec{b}) \cdot (\vec{a}+2\vec{b}) = 20$ $|\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 + |\vec{a}|^2 + 4(\vec{a} \cdot \vec{b}) + 4|\vec{b}|^2 = 20$ $2|\vec{a}|^2 + 5|\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 20$ Substituting the given values: $2(1)^2 + 5(2)^2 + 2(1)(2)\cos\theta = 20$ $2 + 20 + 4\cos\theta = 20$ $22 + 4\cos\theta = 20$ $4\cos\theta = -2$ $\cos\theta = -\frac{1}{2}$ Therefore, $\theta = \frac{2\pi}{3}$.

Class 12 Mathematics Vector Algebra Scalar or Dot product of two vectors and its applications

Easy

If $|\vec{a}|=1, |\vec{b}|=2, |\vec{a}-\vec{b}|^2+|\vec{a}+2\vec{b}|^2=20$, then the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is:

TS EAMCET 2023 All TS EAMCET

A
$\frac{\pi}{3}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{6}$
D
$\frac{2\pi}{3}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Vector Algebra

Subtopic

Scalar or Dot product of two vectors and its applications

Previous Year

TS EAMCET 2023 Paper All TS EAMCET years →

If the moduli of the vectors $a, b, c$ are $3, 4, 5$ respectively and $a$ and $b + c$,$b$ and $c + a$,$c$ and $a + b$ are mutually perpendicular,then the modulus of $a + b + c$ is

MediumIIT 1981

View Solution

In a triangle $ABC,$ if $|\overline{BC}|=8, |\overline{CA}|=7, |\overline{AB}|=10,$ then the projection of the vector $\overline{AB}$ on $\overline{AC}$ is equal to ....... .

MediumJEE MAIN 2021

View Solution

If $|a| = |b|$,then $(a + b) \cdot (a - b)$ is:

Easy

View Solution

If $a \cdot b = a \cdot c$,$a \times b = a \times c$ and $a \neq 0$,then

Easy

View Solution

If $a, b, c$ are the position vectors of the points $A, B, C$ respectively, then match the items of List-$I$ with those of List-$II$.
List-$I$ List-$II$
$A$. $a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = 3\hat{i} + 4\hat{j} + 2\hat{k}, c = 4\hat{i} + 2\hat{j} + 3\hat{k}$ $I$. $\triangle ABC$ is an equilateral triangle
$B$. $a = \hat{i} + 2\hat{j} + 3\hat{k}, b = 3\hat{i} + 4\hat{j} + 7\hat{k}, c = -3\hat{i} - 2\hat{j} - 5\hat{k}$ $II$. $\triangle ABC$ is an isosceles triangle
$C$. $a = 2\hat{i} - \hat{j} + \hat{k}, b = \hat{i} - 3\hat{j} - 5\hat{k}, c = -3\hat{i} - 4\hat{j} - 4\hat{k}$ $III$. $\triangle ABC$ is a right-angled triangle
$D$. $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2\hat{j} + 3\hat{k}, c = 2\hat{i} - \hat{j} + \hat{k}$ $IV$. $A, B, C$ are collinear

The correct match is:

EasyTS EAMCET 2020

View Solution

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

List-$I$	List-$II$
$A$. $a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = 3\hat{i} + 4\hat{j} + 2\hat{k}, c = 4\hat{i} + 2\hat{j} + 3\hat{k}$	$I$. $\triangle ABC$ is an equilateral triangle
$B$. $a = \hat{i} + 2\hat{j} + 3\hat{k}, b = 3\hat{i} + 4\hat{j} + 7\hat{k}, c = -3\hat{i} - 2\hat{j} - 5\hat{k}$	$II$. $\triangle ABC$ is an isosceles triangle
$C$. $a = 2\hat{i} - \hat{j} + \hat{k}, b = \hat{i} - 3\hat{j} - 5\hat{k}, c = -3\hat{i} - 4\hat{j} - 4\hat{k}$	$III$. $\triangle ABC$ is a right-angled triangle
$D$. $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2\hat{j} + 3\hat{k}, c = 2\hat{i} - \hat{j} + \hat{k}$	$IV$. $A, B, C$ are collinear

If $|\vec{a}|=1, |\vec{b}|=2, |\vec{a}-\vec{b}|^2+|\vec{a}+2\vec{b}|^2=20$, then the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is:

Similar Questions

If the moduli of the vectors $a, b, c$ are $3, 4, 5$ respectively and $a$ and $b + c$,$b$ and $c + a$,$c$ and $a + b$ are mutually perpendicular,then the modulus of $a + b + c$ is

In a triangle $ABC,$ if $|\overline{BC}|=8, |\overline{CA}|=7, |\overline{AB}|=10,$ then the projection of the vector $\overline{AB}$ on $\overline{AC}$ is equal to ....... .

If $|a| = |b|$,then $(a + b) \cdot (a - b)$ is:

If $a \cdot b = a \cdot c$,$a \times b = a \times c$ and $a \neq 0$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module