Show that the four points A, B, C and D with the position vectors a , b , c and d respectively are coplanar if and only if 3 a -2 b + c -2 d = 0.

Necessary Condition: Firstly, let a , b , c are coplanar vectors. Then, one of them is expressible as a linear combination of the other two.Let c =x a +y b for some scalars x, y. Then, l a +m b +n c = 0, where l = x, m = y, n = -1. Thus, if a , b , c are coplanar vectors, then there exists a scalars l, m, n not all zero simultaneously satisfying l a +m b +n c = 0 where l, m, n are not all zero simultaneously. Sufficient Condition: Let a , b , c are three scalars such that there exists scalars l, m, n not all zero simultaneously satisfying l a +m b +n c = 0. We have to prove that a , b , c are coplanar vectors. Now, l a +m b +n c = 0 c =-l a -m b c = ( -l n ) a + ( -m n ) b c is a linear combination of a and b . c lies in a plane a and b . Hence, a , b , c are coplanar vectors.

Show that the four points A, B, C and D with the position vectors…

Download App

Question

Show that the four points A, B, C and D with the position vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ and $\vec{\text{d}}$ respectively are coplanar if and only if $3\vec{\text{a}}-2\vec{\text{b}}+\vec{\text{c}}-2\vec{\text{d}}=\vec0$.

✓

Answer

Necessary Condition: Firstly, let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors. Then, one of them is expressible as a linear combination of the other two.Let $\vec{\text{c}}=\text{x}\vec{\text{a}}+\text{y}\vec{\text{b}}$ for some scalars x, y. Then,
$\Rightarrow\ \text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$, where l = x, m = y, n = -1.
Thus, if $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors, then there exists a scalars l, m, n not all zero simultaneously satisfying $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$ where l, m, n are not all zero simultaneously.
Sufficient Condition: Let $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are three scalars such that there exists scalars l, m, n not all zero simultaneously satisfying $\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$. We have to prove that $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors.
Now,
$\text{l}\vec{\text{a}}+\text{m}\vec{\text{b}}+\text{n}\vec{\text{c}}=\vec0$
$\Rightarrow\text{n}\vec{\text{c}}=-\text{l}\vec{\text{a}}-\text{m}\vec{\text{b}}$
$\Rightarrow\ \vec{\text{c}}=\Big(\frac{-\text{l}}{\text{n}}\Big)\vec{\text{a}}+\Big(\frac{-\text{m}}{\text{n}}\Big)\vec{\text{b}}$
$\vec{\text{c}}$ is a linear combination of $\vec{\text{a}}\text{ and }\vec{\text{b}}$.
$\vec{\text{c}}$ lies in a plane $\vec{\text{a}}\text{ and }\vec{\text{b}}$.
Hence, $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are coplanar vectors.