If bar{a}, bar{b}, bar{c} are three vectors s | vedclass

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

Question

(D) Given: $|\bar{a}| = \sqrt{31}$,$|\bar{b}| = \frac{1}{2}$,$|\bar{c}| = 2$,and the angle between $\bar{b}$ and $\bar{c}$ is $	heta = \frac{2\pi}{3}

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(D) Given: $|\bar{a}| = \sqrt{31}$,$|\bar{b}| = \frac{1}{2}$,$|\bar{c}| = 2$,and the angle between $\bar{b}$ and $\bar{c}$ is $	heta = \frac{2\pi}{3}$.From $2(\bar{a} 	imes \bar{b}) = 3(\bar{c} 	imes \bar{a})$,we can write $2(\bar{a} 	imes \bar{b}) + 3(\bar{a} 	imes \bar{c}) = 0$,which implies $\bar{a} 	imes (2\bar{b} + 3\bar{c}) = 0$.This means $\bar{a}$ is parallel to $(2\bar{b} + 3\bar{c})$. Let $2\bar{b} + 3\bar{c} = k\bar{a}$.Squaring both sides: $|2\bar{b} + 3\bar{c}|^2 = k^2|\bar{a}|^2$.$4|\bar{b}|^2 + 9|\bar{c}|^2 + 12(\bar{b} \cdot \bar{c}) = k^2(31)$.$4(\frac{1}{4}) + 9(4) + 12(\frac{1}{2})(2)\cos(\frac{2\pi}{3}) = 31k^2$.$1 + 36 + 12(-1/2) = 31k^2 \implies 31 = 31k^2 \implies k^2 = 1$.We need to find $X = \left|\frac{\bar{a} 	imes \bar{c}}{\bar{a} \cdot \bar{b}}ight|^2$.Since $\bar{a} 	imes (2\bar{b} + 3\bar{c}) = 0$,we have $2(\bar{a} 	imes \bar{b}) = -3(\bar{a} 	imes \bar{c})$,so $\bar{a} 	imes \bar{c} = -\frac{2}{3}(\bar{a} 	imes \bar{b})$.Then $|\bar{a} 	imes \bar{c}|^2 = \frac{4}{9}|\bar{a} 	imes \bar{b}|^2$.Also,$\bar{a} \cdot (2\bar{b} + 3\bar{c}) = k|\bar{a}|^2$. Since $\bar{a} \perp (2\bar{b} + 3\bar{c})$ is not necessarily true,we use the cross product relation: $\bar{a} 	imes \bar{c} = -\frac{2}{3}(\bar{a} 	imes \bar{b})$.Taking magnitudes: $|\bar{a}||\bar{c}|\sin\alpha = \frac{2}{3}|\bar{a}||\bar{b}|\sin\beta$.Using the relation $2\bar{b} + 3\bar{c} = k\bar{a}$,we find the value is $11$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$,then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then,its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$,then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

Similar Questions

If $b$ and $c$ are non-collinear vectors,$|c| \neq 0$,$a \times(b \times c)+(a \cdot b) b=(4-2 \beta-\sin \alpha) b+\left(\beta^2-1\right) c$ and $(c \cdot c) a=c$,then the scalars $\alpha$ and $\beta$ are

The minimum value of $(x_1 - x_2)^2 + (\sqrt{2 - x_1^2} - \frac{9}{x_2})^2$ where $x_1 \in (0, \sqrt{2})$ and $x_2 \in R^+$.

If $ABCD$ is a quadrilateral,then the resultant force represented by $\vec{BA}, \vec{BC}, \vec{CD}$ and $\vec{DA}$ is equal to:

If the vector $\vec{a} = (x, y, z)$ makes an obtuse angle with the $y$-axis and makes equal angles with the vectors $\vec{b} = (y, -2z, 3x)$ and $\vec{c} = (2z, 3x, -y)$,and if $|\vec{a}| = 2\sqrt{3}$ and $\vec{a}$ is perpendicular to $\vec{d} = (1, -1, 2)$,find the vector $\vec{a}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module