(C) The magnitude of a vector remains invariant under the rotation of the coordinate system about the origin.Given the components of vector $\bar{a}$

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

(C) The magnitude of a vector remains invariant under the rotation of the coordinate system about the origin.Given the components of vector $\bar{a}$ in the original system are $(1, 2p)$,its squared magnitude is:$|\bar{a}|^2 = 1^2 + (2p)^2 = 1 + 4p^2$In the new system,the components are $(1, p+1)$,so its squared magnitude is:$|\bar{b}|^2 = 1^2 + (p+1)^2 = 1 + p^2 + 2p + 1 = p^2 + 2p + 2$Since $|\bar{a}|^2 = |\bar{b}|^2$,we have:$1 + 4p^2 = p^2 + 2p + 2$$3p^2 - 2p - 1 = 0$Factoring the quadratic equation:$(3p + 1)(p - 1) = 0$Thus,$p = -\frac{1}{3}$ or $p = 1$.

Class 12 Mathematics Vector Algebra Basic , Modulus and Algebra of vectors

Difficult

$A$ vector $\bar{a}$ has components $1$ and $2p$ with respect to a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If,with respect to the new system,$\bar{a}$ has components $1$ and $(p+1)$,then:

MHT CET 2023 All MHT CET

A
$p=1$ or $p=\frac{1}{3}$
B
$p=-1$ or $p=\frac{-1}{3}$
C
$p=\frac{-1}{3}$ or $p=1$
D
$p=\frac{1}{3}$ or $p=-1$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Vector Algebra

Subtopic

Basic , Modulus and Algebra of vectors

Previous Year

MHT CET 2023 Paper All MHT CET years →

If $m_1, m_2, m_3$ and $m_4$ are respectively the magnitudes of the vectors $\overrightarrow{a}_1=2 \hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{a}_2=3 \hat{i}-4 \hat{j}-4 \hat{k}$, $\overrightarrow{a}_3=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{a}_4=-\hat{i}+3 \hat{j}+\hat{k}$, then the correct order of $m_1, m_2, m_3$ and $m_4$ is

DifficultAP EAMCET 2009

View Solution

If $a = 2i + 5j$ and $b = 2i - j,$ then the unit vector along $a + b$ will be

Easy

View Solution

Find the values of $x, y$ and $z$ so that the vectors $\vec{a} = x \hat{i} + 2 \hat{j} + z \hat{k}$ and $\vec{b} = 2 \hat{i} + y \hat{j} + \hat{k}$ are equal.

Easy

View Solution

If $ABCD$ is a parallelogram,$\overrightarrow{AB} = 2i + 4j - 5k$ and $\overrightarrow{AD} = i + 2j + 3k,$ then the unit vector in the direction of $\overrightarrow{BD}$ is

Medium

View Solution

If $P \hat{i}-2 \hat{j}+3 \hat{k}$, $2 \hat{i}+3 \hat{j}-4 \hat{k}$, and $4 \hat{i}+13 \hat{j}-18 \hat{k}$ are the position vectors of three collinear points $A$, $B$, and $C$ respectively, then the vector in the direction of $AB$ of length $|P|$ units is

EasyTS EAMCET 2021

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

$A$ vector $\bar{a}$ has components $1$ and $2p$ with respect to a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If,with respect to the new system,$\bar{a}$ has components $1$ and $(p+1)$,then:

Similar Questions

If $a = 2i + 5j$ and $b = 2i - j,$ then the unit vector along $a + b$ will be

Find the values of $x, y$ and $z$ so that the vectors $\vec{a} = x \hat{i} + 2 \hat{j} + z \hat{k}$ and $\vec{b} = 2 \hat{i} + y \hat{j} + \hat{k}$ are equal.

If $ABCD$ is a parallelogram,$\overrightarrow{AB} = 2i + 4j - 5k$ and $\overrightarrow{AD} = i + 2j + 3k,$ then the unit vector in the direction of $\overrightarrow{BD}$ is

If $P \hat{i}-2 \hat{j}+3 \hat{k}$, $2 \hat{i}+3 \hat{j}-4 \hat{k}$, and $4 \hat{i}+13 \hat{j}-18 \hat{k}$ are the position vectors of three collinear points $A$, $B$, and $C$ respectively, then the vector in the direction of $AB$ of length $|P|$ units is

Vedclass Test Series

Exam Paper Generator

Online Exam Module