Let overline{OA} = vec{a},overline{OB} = 10ve | vedclass

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

Question

(B) The area of the quadrilateral $OABC$ can be calculated as the sum of the areas of $	riangle OAB$ and $	riangle OBC$.Area of $	riangle OAB = \fr

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The area of the quadrilateral $OABC$ can be calculated as the sum of the areas of $	riangle OAB$ and $	riangle OBC$.Area of $	riangle OAB = \frac{1}{2} |\overline{OA} 	imes \overline{OB}| = \frac{1}{2} |\vec{a} 	imes (10\vec{a} + 2\vec{b})| = \frac{1}{2} |10(\vec{a} 	imes \vec{a}) + 2(\vec{a} 	imes \vec{b})| = \frac{1}{2} |0 + 2(\vec{a} 	imes \vec{b})| = |\vec{a} 	imes \vec{b}|$.Area of $	riangle OBC = \frac{1}{2} |\overline{OB} 	imes \overline{OC}| = \frac{1}{2} |(10\vec{a} + 2\vec{b}) 	imes \vec{b}| = \frac{1}{2} |10(\vec{a} 	imes \vec{b}) + 2(\vec{b} 	imes \vec{b})| = \frac{1}{2} |10(\vec{a} 	imes \vec{b}) + 0| = 5|\vec{a} 	imes \vec{b}|$.Thus,$p = |\vec{a} 	imes \vec{b}| + 5|\vec{a} 	imes \vec{b}| = 6|\vec{a} 	imes \vec{b}|$.The area $q$ of the parallelogram with adjacent sides $OA$ and $OC$ is given by $q = |\vec{a} 	imes \vec{b}|$.Therefore,$p/q = \frac{6|\vec{a} 	imes \vec{b}|}{|\vec{a} 	imes \vec{b}|} = 6$.

Let $\overline{OA} = \vec{a}$,$\overline{OB} = 10\vec{a} + 2\vec{b}$,and $\overline{OC} = \vec{b}$,where $O, A, C$ are non-collinear. Let $p$ be the area of the quadrilateral $OABC$ and $q$ be the area of the parallelogram with adjacent sides $OA$ and $OC$. Then $p/q = \dots$

Similar Questions

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\pi / 3$,then $\vec{a}$ is equal to

If $a = 2i + 2j - k$ and $b = 6i - 3j + 2k$,then what is the value of $a \times b$?

If $A, B, C, D$ are four points in space,then $|\overline{AB} \times \overline{CD} + \overline{BC} \times \overline{AD} + \overline{CA} \times \overline{BD}| = \lambda \times (\text{Area of } \Delta ABC)$. Find $\lambda$.

If the two diagonals of a parallelogram are $\bar{d_1} = \bar{i} + 2\bar{j} + 3\bar{k}$ and $\bar{d_2} = -2\bar{i} + \bar{j} - 2\bar{k}$,then the area of the parallelogram in square units is

Suppose $L_1$ and $L_2$ are two lines having the direction ratios $1, -2, -2$ and $0, 2, 1$ respectively. If the direction cosines of a line perpendicular to both $L_1$ and $L_2$ are $l, m, n$, then $|l| + |m| + |n| =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module