If A, B, C and D are four points in the plane | vedclass

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

Question

(B) Let the position vectors of points $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively. Given that $|\vec{b}-\vec{a}|^2 + |\vec{d}-\

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Let the position vectors of points $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively. Given that $|\vec{b}-\vec{a}|^2 + |\vec{d}-\vec{c}|^2 = |\vec{c}-\vec{b}|^2 + |\vec{a}-\vec{d}|^2$. Expanding the squares using the dot product property $|\vec{x}-\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$: $|\vec{b}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{d}|^2 + |\vec{c}|^2 - 2\vec{c} \cdot \vec{d} = |\vec{c}|^2 + |\vec{b}|^2 - 2\vec{b} \cdot \vec{c} + |\vec{a}|^2 + |\vec{d}|^2 - 2\vec{d} \cdot \vec{a}$. Canceling common terms $|\vec{a}|^2, |\vec{b}|^2, |\vec{c}|^2, |\vec{d}|^2$ from both sides: $-2\vec{a} \cdot \vec{b} - 2\vec{c} \cdot \vec{d} = -2\vec{b} \cdot \vec{c} - 2\vec{d} \cdot \vec{a}$. Rearranging the terms: $\vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{d} - \vec{b} \cdot \vec{c} - \vec{d} \cdot \vec{a} = 0$. Factorizing the expression: $(\vec{a}-\vec{c}) \cdot (\vec{b}-\vec{d}) = 0$. This implies $\vec{AC} \cdot \vec{DB} = 0$,which means $\vec{AC} \cdot \vec{BD} = 0$.

If $A, B, C$ and $D$ are four points in the plane such that $|AB|^2+|CD|^2=|BC|^2+|DA|^2=100$,then $AC \cdot BD=$

Similar Questions

Let $u, v, w$ be three vectors such that $|u| = 1, |v| = 2, |w| = 3$. If the projection of $v$ along $u$ is equal to the projection of $w$ along $u$ and $v, w$ are perpendicular to each other, then $|u - v + w| = ...$

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$,and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are respectively the position vectors of the vertices $A, B, C$ of $\triangle ABC$,then the position vector of the point where the bisector of angle $A$ meets $BC$ is

If $\bar{a}, \bar{b}$ and $\bar{c}$ are vectors such that $|\bar{a}| = |\frac{\bar{b}}{2}| = |\frac{\bar{c}}{3}| = 1$; $\bar{b}$ and $\bar{c}$ are perpendicular; and the projections of $\bar{b}$ and $\bar{c}$ on $\bar{a}$ are equal,then $|\bar{a} - \bar{b} + \bar{c}| = $

If $\vec{a}$ and $\vec{b}$ are unit vectors such that the vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$,then find the angle between $\vec{a}$ and $\vec{b}$.

If $a=\hat{i}+\hat{j}+t \hat{k}$ and $b=\hat{i}+2 \hat{j}+3 \hat{k}$, then the values of $t$ for which $(a+b)$ and $(a-b)$ are perpendicular are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module