In a trapezium,the vector overrightarrow{BC} | vedclass

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

Question

(A) Given that $\overrightarrow{BC} = \lambda \overrightarrow{AD}$.We have $p = \overrightarrow{AC} + \overrightarrow{BD}$.Using vector addition,$\ove

Vedclass · Accepted Answer

$\mu = \lambda + 1$

Vedclass · Answer

(A) Given that $\overrightarrow{BC} = \lambda \overrightarrow{AD}$.We have $p = \overrightarrow{AC} + \overrightarrow{BD}$.Using vector addition,$\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD}$.Substituting this into the expression for $p$:$p = \overrightarrow{AC} + (\overrightarrow{BC} + \overrightarrow{CD}) = (\overrightarrow{AC} + \overrightarrow{CD}) + \overrightarrow{BC}$.Since $\overrightarrow{AC} + \overrightarrow{CD} = \overrightarrow{AD}$,we get:$p = \overrightarrow{AD} + \overrightarrow{BC}$.Substituting $\overrightarrow{BC} = \lambda \overrightarrow{AD}$:$p = \overrightarrow{AD} + \lambda \overrightarrow{AD} = (1 + \lambda) \overrightarrow{AD}$.Given $p = \mu \overrightarrow{AD}$,by comparing the coefficients,we find $\mu = \lambda + 1$.

In a trapezium,the vector $\overrightarrow{BC} = \lambda \overrightarrow{AD}$. We will then find that $p = \overrightarrow{AC} + \overrightarrow{BD}$ is collinear with $\overrightarrow{AD}$. If $p = \mu \overrightarrow{AD}$,then

Similar Questions

The position vectors of the points $A$ and $B$ with respect to $O$ are $2 \hat{i}+2 \hat{j}+\hat{k}$ and $2 \hat{i}+4 \hat{j}+4 \hat{k}$. The length of the internal bisector of $\angle BOA$ of $\triangle AOB$ is:

If $|\vec{a}|=16$ and $|\vec{b}|=4$,then find the value of $\sqrt{|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}}$.

Two adjacent sides of a triangle are represented by the vectors $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = 2\sqrt{3}\hat{i} - 2\sqrt{3}\hat{j} + \sqrt{3}\hat{k}$. Then the least angle of the triangle and the perimeter of the triangle are respectively:

If the position vectors of the vertices of a triangle are $2 \hat{i}-\hat{j}+\hat{k}$,$\hat{i}-3 \hat{j}-5 \hat{k}$,and $3 \hat{i}-4 \hat{j}-4 \hat{k}$,then the triangle is

If $\overrightarrow{a}=\hat{i}+3 \hat{j}+13 \hat{k}$ and $\overrightarrow{b}=2 \hat{i}-4 \hat{j}+3 \hat{k}$ are two vectors,then the component vector of $\vec{a}$ perpendicular to $\vec{b}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module