If |overrightarrow{a}|=2, |overrightarrow{b}| | vedclass

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

Question

(C) Let the vertices of the triangle be $O(0), A(\overrightarrow{a}+\overrightarrow{b}),$ and $B(\overrightarrow{a}-\overrightarrow{b})$.The area of t

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Let the vertices of the triangle be $O(0), A(\overrightarrow{a}+\overrightarrow{b}),$ and $B(\overrightarrow{a}-\overrightarrow{b})$.The area of the triangle is given by $\frac{1}{2} |\overrightarrow{OA} 	imes \overrightarrow{OB}|$.Substituting the vectors,we get $	ext{Area} = \frac{1}{2} |(\overrightarrow{a}+\overrightarrow{b}) 	imes (\overrightarrow{a}-\overrightarrow{b})|$.Expanding the cross product: $(\overrightarrow{a}+\overrightarrow{b}) 	imes (\overrightarrow{a}-\overrightarrow{b}) = \overrightarrow{a} 	imes \overrightarrow{a} - \overrightarrow{a} 	imes \overrightarrow{b} + \overrightarrow{b} 	imes \overrightarrow{a} - \overrightarrow{b} 	imes \overrightarrow{b}$.Since $\overrightarrow{a} 	imes \overrightarrow{a} = 0$ and $\overrightarrow{b} 	imes \overrightarrow{b} = 0$,and $\overrightarrow{b} 	imes \overrightarrow{a} = -(\overrightarrow{a} 	imes \overrightarrow{b})$,we have:$= 0 - (\overrightarrow{a} 	imes \overrightarrow{b}) - (\overrightarrow{a} 	imes \overrightarrow{b}) - 0 = -2(\overrightarrow{a} 	imes \overrightarrow{b})$.Thus,$	ext{Area} = \frac{1}{2} |-2(\overrightarrow{a} 	imes \overrightarrow{b})| = |\overrightarrow{a} 	imes \overrightarrow{b}|$.Since $\overrightarrow{a} \perp \overrightarrow{b}$,$|\overrightarrow{a} 	imes \overrightarrow{b}| = |\overrightarrow{a}| |\overrightarrow{b}| \sin(90^{\circ}) = 2 	imes 3 	imes 1 = 6$.

If $|\overrightarrow{a}|=2, |\overrightarrow{b}|=3$ and $\overrightarrow{a}, \overrightarrow{b}$ are mutually perpendicular,then the area of the triangle whose vertices are $\overrightarrow{0}, \overrightarrow{a}+\overrightarrow{b}, \overrightarrow{a}-\overrightarrow{b}$ is

Similar Questions

The locus of the point $P(\vec{r})$ which forms a triangle $ABP$ of area $1$ sq. unit with the fixed points $A(\hat{i})$ and $B(\hat{j})$ is

Let $A(2, 3, 5)$ and $C(-3, 4, -2)$ be opposite vertices of a parallelogram $ABCD$. If the diagonal $\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$,then the area of the parallelogram is equal to:

Let $\overrightarrow{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=((\overrightarrow{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is:

Let $\hat{\alpha}, \hat{\beta}, \hat{\gamma}$ be three unit vectors such that $\hat{\alpha} \times (\hat{\beta} \times \hat{\gamma}) = \frac{1}{2}(\hat{\beta} + \hat{\gamma})$. If $\hat{\beta}$ is not parallel to $\hat{\gamma}$,then the angle between $\hat{\alpha}$ and $\hat{\beta}$ is

If $1, 2, 3$ and $-1, 0, 1$ are the direction ratios of the rays $OA$ and $OB$ respectively,then the direction cosines of a normal to the plane $AOB$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module