Let overrightarrow{OA}=2 overrightarrow{a},ov | vedclass

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

Question

(D) The area of a parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is given by $|\overrightarrow{OA} 	imes \overrigh

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(D) The area of a parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is given by $|\overrightarrow{OA} 	imes \overrightarrow{OC}|$.Given,$|\overrightarrow{OA} 	imes \overrightarrow{OC}| = |2 \overrightarrow{a} 	imes 3 \overrightarrow{b}| = 15$.$6 |\overrightarrow{a} 	imes \overrightarrow{b}| = 15 \implies |\overrightarrow{a} 	imes \overrightarrow{b}| = \frac{15}{6} = \frac{5}{2} \dots (1)$.The area of a quadrilateral $OABC$ with diagonals $\overrightarrow{OB}$ and $\overrightarrow{AC}$ is given by $\frac{1}{2} |\overrightarrow{OB} 	imes \overrightarrow{AC}|$.We have $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = 3 \overrightarrow{b} - 2 \overrightarrow{a}$.Area $= \frac{1}{2} |(6 \overrightarrow{a} + 5 \overrightarrow{b}) 	imes (3 \overrightarrow{b} - 2 \overrightarrow{a})|$.$= \frac{1}{2} |18 (\overrightarrow{a} 	imes \overrightarrow{b}) - 12 (\overrightarrow{a} 	imes \overrightarrow{a}) + 15 (\overrightarrow{b} 	imes \overrightarrow{b}) - 10 (\overrightarrow{b} 	imes \overrightarrow{a})|$.Since $\overrightarrow{a} 	imes \overrightarrow{a} = 0$,$\overrightarrow{b} 	imes \overrightarrow{b} = 0$,and $\overrightarrow{b} 	imes \overrightarrow{a} = -(\overrightarrow{a} 	imes \overrightarrow{b})$:Area $= \frac{1}{2} |18 (\overrightarrow{a} 	imes \overrightarrow{b}) + 10 (\overrightarrow{a} 	imes \overrightarrow{b})| = \frac{1}{2} |28 (\overrightarrow{a} 	imes \overrightarrow{b})| = 14 |\overrightarrow{a} 	imes \overrightarrow{b}|$.Substituting from $(1)$:Area $= 14 	imes \frac{5}{2} = 35$ sq. units.

Similar Questions

Find a unit vector coplanar with $i + 2j + k$ and $i + j + 2k$ and perpendicular to $i + j + k$.

If $a = i + j + k$,$b = i + 3j + 5k$,and $c = 7i + 9j + 11k$,then the area of the parallelogram having diagonals $a + b$ and $b + c$ is

Let $\bar{a}=4 \bar{i}+5 \bar{j}-\bar{k}$,$\bar{b}=\bar{i}-4 \bar{j}+5 \bar{k}$,$\bar{c}=3 \bar{i}+\bar{j}-\bar{k}$ and let $\bar{\alpha}$ be a vector perpendicular to both $\bar{a}$ and $\bar{b}$ such that $\bar{\alpha} \cdot \bar{c}=63$. Then $\bar{\alpha}=$

If $\vec{a}=\hat{i}+\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ is a unit vector perpendicular to $\vec{a}$ and coplanar with $\vec{a}$ and $\vec{b}$, then the unit vector $\vec{d}$ perpendicular to both $\vec{a}$ and $\vec{c}$ is

Find the unit vector perpendicular to the plane containing the vectors $\vec{a} = 2\hat{i} - 6\hat{j} - 3\hat{k}$ and $\vec{b} = 4\hat{i} + 3\hat{j} - \hat{k}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module