If $A, B, C, D$ are the points with position vectors $\hat{i}-\hat{j}+\hat{k}, 2 \hat{i}-\hat{j}+3 \hat{k}, 2 \hat{i}-3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$ respectively,find the projection of $\overrightarrow{AB}$ on $\overrightarrow{CD}$.
(D) Given position vectors are $\overrightarrow{OA} = \hat{i}-\hat{j}+\hat{k}$,$\overrightarrow{OB} = 2\hat{i}-\hat{j}+3\hat{k}$,$\overrightarrow{OC} = 2\hat{i}-3\hat{k}$,and $\overrightarrow{OD} = 3\hat{i}-2\hat{j}+\hat{k}$.
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2-1)\hat{i} + (-1 - (-1))\hat{j} + (3-1)\hat{k} = \hat{i} + 0\hat{j} + 2\hat{k}$.
$\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = (3-2)\hat{i} + (-2-0)\hat{j} + (1 - (-3))\hat{k} = \hat{i} - 2\hat{j} + 4\hat{k}$.
The projection of $\overrightarrow{AB}$ on $\overrightarrow{CD}$ is given by $\frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{CD}|}$.
$\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(1) + (0)(-2) + (2)(4) = 1 + 0 + 8 = 9$.
$|\overrightarrow{CD}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.
Therefore,the projection is $\frac{9}{\sqrt{21}} = \frac{9\sqrt{21}}{21} = \frac{3\sqrt{21}}{7}$ units.
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2-1)\hat{i} + (-1 - (-1))\hat{j} + (3-1)\hat{k} = \hat{i} + 0\hat{j} + 2\hat{k}$.
$\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = (3-2)\hat{i} + (-2-0)\hat{j} + (1 - (-3))\hat{k} = \hat{i} - 2\hat{j} + 4\hat{k}$.
The projection of $\overrightarrow{AB}$ on $\overrightarrow{CD}$ is given by $\frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{CD}|}$.
$\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(1) + (0)(-2) + (2)(4) = 1 + 0 + 8 = 9$.
$|\overrightarrow{CD}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.
Therefore,the projection is $\frac{9}{\sqrt{21}} = \frac{9\sqrt{21}}{21} = \frac{3\sqrt{21}}{7}$ units.
Explore More
Similar Questions
Two adjacent sides of a parallelogram $ABCD$ are given by $\overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $AD$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $AD$ becomes $AD'$. If $AD'$ makes a right angle with the side $AB$,then the cosine of the angle $\alpha$ is given by
DifficultIIT 2010
View SolutionIf the vertices $A, B, C$ of a triangle $ABC$ are $(1,2,3), (-1,0,0), (0,1,2)$ respectively,then find $\angle ABC$. $[\angle ABC \text{ is the angle between the vectors } \overrightarrow{BA} \text{ and } \overrightarrow{BC}]$.
Medium
View SolutionIt is given that $a, b, c$ are vectors of lengths $6, 8, 10$ respectively. If $a$ is perpendicular to $(b+c)$, $b$ is perpendicular to $(c+a)$, and $c$ is perpendicular to $(a+b)$, then the length of the vector $a+b+c$ is (in $\sqrt{2}$)
Let $\vec{u} = 2\hat{i} + 3\hat{j} + \hat{k}$,$\vec{v} = -3\hat{i} + 2\hat{j}$ and $\vec{w} = \hat{i} - \hat{j} + 4\hat{k}$. Then which of the following statements is true?
Let two non-collinear vectors $\hat{a}$ and $\hat{b}$ form an acute angle. $A$ point $P$ moves such that at any time $t$,the position vector $\overline{OP}$,where $O$ is the origin,is given by $\hat{a} \sin t + \hat{b} \cos t$. When $P$ is farthest from the origin $O$,let $M$ be the length of $\overline{OP}$ and $\hat{u}$ be the unit vector along $\overline{OP}$. Then:
DifficultMHT CET 2023
View SolutionVedclass 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