Given that $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \times \vec{b} = \vec{0}$. What can you conclude about the vectors $\vec{a}$ and $\vec{b}$?
(A) Given $\vec{a} \cdot \vec{b} = 0$. This implies that either $|\vec{a}| = 0$ or $|\vec{b}| = 0$,or $\vec{a} \perp \vec{b}$ (if both are non-zero vectors).
Given $\vec{a} \times \vec{b} = \vec{0}$. This implies that either $|\vec{a}| = 0$ or $|\vec{b}| = 0$,or $\vec{a} \parallel \vec{b}$ (if both are non-zero vectors).
Since two non-zero vectors cannot be both perpendicular and parallel to each other simultaneously,the only possibility is that at least one of the vectors must be a zero vector.
Therefore,we conclude that either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$.
Given $\vec{a} \times \vec{b} = \vec{0}$. This implies that either $|\vec{a}| = 0$ or $|\vec{b}| = 0$,or $\vec{a} \parallel \vec{b}$ (if both are non-zero vectors).
Since two non-zero vectors cannot be both perpendicular and parallel to each other simultaneously,the only possibility is that at least one of the vectors must be a zero vector.
Therefore,we conclude that either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$.
Explore More
Similar Questions
If $\overrightarrow{AB} = 2\hat{i} + 3\hat{j} - 6\hat{k}$ and $\overrightarrow{BC} = 6\hat{i} - 2\hat{j} + 3\hat{k}$ are the vectors along two sides of a triangle $ABC$,then the perimeter of triangle $ABC$ is:
Three forces of magnitudes $1, 2, 3 \text{ dynes}$ meet at a point and act along the diagonals of three adjacent faces of a cube. The resultant force is ............ $\text{dyne}$.
Difficult
View SolutionCompute the magnitude of the following vectors:
$\vec{a}=\hat{i}+\hat{j}+\hat{k} ; \quad \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k} ; \quad \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$
$\vec{a}=\hat{i}+\hat{j}+\hat{k} ; \quad \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k} ; \quad \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$
Easy
View SolutionLet $ABC$ be a triangle and $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of $A, B, C$ respectively. If $D$ divides $BC$ in the ratio $2:3$ internally and $E$ divides $CA$ in the ratio $2:1$ internally, then the position vector of the point $P$ which divides $DE$ in the ratio $3:5$ internally is
The magnitudes of mutually perpendicular forces $a, b$ and $c$ are $2, 10$ and $11$ respectively. Then the magnitude of its resultant is
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo