MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Test for Coplanarity

Four points g...

Question

Four points given by position vectors $2 \to a + 3 \to b - \to c, \to a - 2 \to b + 3 \to c, 3 \to a + 4 \to b - 2 \to c a n d \to a - 6 \to b + 6 \to c$ are coplanar, where $\to a, \to b a n d \to c$ are non-coplanar vectors.

A

True

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A True
Let four points be A, B, C and D respectively and O be the origin, then
$\begin{matrix} O \to A = 2 \to a + 3 \to b - \to c O \to B = \to a - 2 \to b + 3 \to c O \to C = 3 \to a + 4 \to b - 2 \to c O \to D = \to a - 6 \to b + 6 \to c \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$ Just because these 4 points are position vectors with respect to origin.
Now let us form 3 vectors out of these 4 points. Why? Because for 3 vectors we know the working rule. We just have to check their linear dependence and if they are linearly dependent, then they are coplanar. Else not.
Let $α = A \to B = O \to B - O \to A$ (Triangle law for vector addition)
$= - \to a - 5 \to b + 4 \to c$
$β = A \to C = O \to C - O \to A$
$= \to a + \to b + \to c$
$γ = A \to D = O \to D - O \to A = - \to a - 9 \to b + 7 \to c$
For linear dependence
Let $\to α = x \to β + y \to γ$
$- \to a - 5 \to b + 4 \to c = x (\to a + \to b + \to c) + y (- \to a - 9 \to b + 7 \to c)$
$= (x - y) \to a + (x - 9 y) \to b - (x - 7 y) \to c$
Equating the coefficients of $\to a, \to b a n d \to c .$ We get
x – y = -1 …(1)
x – 9y = -5 …(2)
-x + 7y = 4 …(3)
Solving (1) and (2) we get $y = \frac{1}{2}$
$x = - \frac{1}{2}$
and these values satisfy equation 3 as well. So system of equations is consistent with a non-zero solution. So $α, β a n d γ$ are coplanar. Hence points A, B, C and D are coplanar. (Because $- - \to A B, - - \to A C a n d - - \to A D$ lie in one plane, so $- - \to O A, - - \to O B - - \to O C a n d - - \to O D$ all must lie in that plane).

Suggest Corrections

thumbs-up

0

Similar questions

Q. Four points given by position vectors

2 \to a + 3 \to b - \to c, \to a - 2 \to b + 3 \to c, 3 \to a + 4 \to b - 2 \to c a n d \to a - 6 \to b + 6 \to c

are coplanar, where

\to a, \to b a n d \to c

are non-coplanar vectors.

\to a, \to b, \to c

are three non coplanar vectors,

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}

\to q = \frac{\to c \times \to a}{[\to a \to b \to c]}

\to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

(2 \to a + 3 \to b + 4 \to c) . \to p + (2 \to b + 3 \to c + 4 \to a) . \to q +

(2 \to c + 3 \to a + 4 \to b) . \to r =

\to a, \to b, \to c

are non-coplanar vectors, then show that the four points

2 \to a + \to b, \to a + 2 \to b + \to c, 4 \to a - 2 \to b - \to c

3 \to a + 4 \to b - 5 \to c

Q. Examine whether following vectors are coplanar or not.

\to a + \to b - 2 \to c, \to a - 3 \to b + \to c

2 \to a - \to b - \to c

\to a, \to b, \to c

be any three non-zero, non-coplanar vectors, then find the linear relation between the following four vectors

\to p = \to a - 2 \to b + 3 \to c

\to q = 2 \to a - 3 \to b + 4 \to c

\to r = 3 \to a - 4 \to b + 5 \to c

\to s = 7 \to a - 11 \to b + 15 \to c .

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Application of Vectors - Planes

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Test for Coplanarity

Standard IX Mathematics

Join BYJU'S Learning Program

CrossIcon