If four distinct points with position vectors | vedclass

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

Question

(D) Since the four points with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar,the vectors $(\vec{b}-\vec{a}), (\vec{c}-\vec{a}),$

Vedclass · Accepted Answer

$[\vec{b} \vec{c} \vec{d}]+[\vec{d} \vec{a} \vec{c}]+[\vec{d} \vec{b} \vec{a}]$

Vedclass · Answer

(D) Since the four points with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar,the vectors $(\vec{b}-\vec{a}), (\vec{c}-\vec{a}),$ and $(\vec{d}-\vec{a})$ are coplanar.Therefore,their scalar triple product is zero: $[\vec{b}-\vec{a}, \vec{c}-\vec{a}, \vec{d}-\vec{a}] = 0$.Expanding this,we get $(\vec{b}-\vec{a}) \cdot ((\vec{c}-\vec{a}) 	imes (\vec{d}-\vec{a})) = 0$.Using the properties of the scalar triple product,this expands to $[\vec{b} \vec{c} \vec{d}] - [\vec{b} \vec{c} \vec{a}] - [\vec{b} \vec{a} \vec{d}] - [\vec{a} \vec{c} \vec{d}] = 0$.Rearranging the terms,we get $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{d}] + [\vec{d} \vec{a} \vec{c}] + [\vec{d} \vec{b} \vec{a}]$.Comparing this with the given options,option $D$ is correct.

If four distinct points with position vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are coplanar,then $[\vec{a} \vec{b} \vec{c}]$ is equal to

Similar Questions

Let $\vec V = 2\hat i + \hat j - \hat k$,$\vec W = \hat i + 3\hat k$,and $|\vec U| = 2$. If $\vec U$ is a vector in the $x-y$ plane,then the greatest value of $([\vec U \vec V \vec W])^2$ is:

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors,then $\frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{\vec{c} \cdot (\vec{a} \times \vec{b})} + \frac{\vec{b} \cdot (\vec{a} \times \vec{c})}{\vec{c} \cdot (\vec{a} \times \vec{b})} = \dots$

If the volume of a tetrahedron having $\bar{i}+2 \bar{j}-3 \bar{k}$,$2 \bar{i}+\bar{j}-3 \bar{k}$,and $3 \bar{i}-\bar{j}+p \bar{k}$ as its coterminous edges is $2$,then the values of $p$ are the roots of the equation

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=2\hat{i}+\lambda\hat{j}+\hat{k}$,$\vec{c}=\hat{i}-\hat{j}+4\hat{k}$ and $\vec{a} \cdot (\vec{b} \times \vec{c}) = 10$,then $\lambda$ is equal to

If $\vec{a}=\alpha \hat{i}+\beta \hat{j}+3 \hat{k}$,$\vec{b}=\hat{j}+2 \hat{k}$,and $\vec{c}=3 \hat{i}+2 \hat{j}+\hat{k}$ are linearly dependent vectors and the magnitude of $\vec{a}$ is $\sqrt{14}$. If $\alpha$ and $\beta$ are integers,then $\alpha+\beta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module