Difficult
If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k},$ find a vector $\vec{c}$ such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3.$
(N/A) Let $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}.$
Given $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k}.$
For $\vec{a} \times \vec{c}=\vec{b},$
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = 0\hat{i} + 1\hat{j} - 1\hat{k}.$
Expanding the determinant,we get $\hat{i}(z-y) - \hat{j}(z-x) + \hat{k}(y-x) = 0\hat{i} + 1\hat{j} - 1\hat{k}.$
Comparing coefficients,we have:
$z-y=0 \implies z=y \quad (i)$
$x-z=1 \implies x=z+1 \quad (ii)$
$y-x=-1 \quad (iii)$
Also,$\vec{a} \cdot \vec{c}=3 \implies x+y+z=3 \quad (iv).$
Substituting $x=z+1$ and $y=z$ into $(iv)$:
$(z+1) + z + z = 3 \implies 3z+1=3 \implies 3z=2 \implies z=\frac{2}{3}.$
Thus,$y=\frac{2}{3}$ and $x=\frac{2}{3}+1=\frac{5}{3}.$
Therefore,$\vec{c}=\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k} = \frac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k}).$
Given $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k}.$
For $\vec{a} \times \vec{c}=\vec{b},$
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = 0\hat{i} + 1\hat{j} - 1\hat{k}.$
Expanding the determinant,we get $\hat{i}(z-y) - \hat{j}(z-x) + \hat{k}(y-x) = 0\hat{i} + 1\hat{j} - 1\hat{k}.$
Comparing coefficients,we have:
$z-y=0 \implies z=y \quad (i)$
$x-z=1 \implies x=z+1 \quad (ii)$
$y-x=-1 \quad (iii)$
Also,$\vec{a} \cdot \vec{c}=3 \implies x+y+z=3 \quad (iv).$
Substituting $x=z+1$ and $y=z$ into $(iv)$:
$(z+1) + z + z = 3 \implies 3z+1=3 \implies 3z=2 \implies z=\frac{2}{3}.$
Thus,$y=\frac{2}{3}$ and $x=\frac{2}{3}+1=\frac{5}{3}.$
Therefore,$\vec{c}=\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k} = \frac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k}).$
Explore More
Similar Questions
Let $a = 2i + j - 2k$ and $b = i + j$. If $c$ is a vector such that $a \cdot c = |c|$,$|c - a| = 2\sqrt{2}$,and the angle between $(a \times b)$ and $c$ is $30^{\circ}$,then $|(a \times b) \times c| = \dots$
Difficult
View Solution$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane defined by $\hat{i}$ and $\hat{i} + \hat{j}$,and the plane defined by $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$. Find the angle between $\vec{a}$ and $\hat{i} - 2\hat{j} + 2\hat{k}$.
Difficult
View SolutionLet the vectors $\overline{PQ}, \overline{QR}, \overline{RS}, \overline{ST}, \overline{TU}$ and $\overline{UP}$ represent the sides of a regular hexagon.
$STATEMENT-1$: $\overline{PQ} \times (\overline{RS} + \overline{ST}) \neq \overrightarrow{0}$.
$STATEMENT-2$: $\overline{PQ} \times \overline{RS} = \overrightarrow{0}$ and $\overline{PQ} \times \overline{ST} \neq \overrightarrow{0}$.
$STATEMENT-1$: $\overline{PQ} \times (\overline{RS} + \overline{ST}) \neq \overrightarrow{0}$.
$STATEMENT-2$: $\overline{PQ} \times \overline{RS} = \overrightarrow{0}$ and $\overline{PQ} \times \overline{ST} \neq \overrightarrow{0}$.
DifficultIIT 2007
View Solution$A$ tetrahedron has vertices $P(1, 2, 1)$,$Q(2, 1, 3)$,$R(-1, 1, 2)$ and $O(0, 0, 0)$. The angle between the faces $OPQ$ and $PQR$ is
DifficultJEE MAIN 2019
View SolutionIf $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$ represent the adjacent sides of a parallelogram,then the area of this parallelogram is:
Easy
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