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Collinear Vectors

If a,b,c ar...

Question

If $a, b, c$ are non-coplanar vectors, then which of the following points are collinear whose position vectors are given by :

a - 2 b + 3 c, 2 a + 3 b - 4 c, - 7 b + 10 c

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3 a - 4 b + 3 c, - 4 a + 5 b - 6 c, 4 a - 7 b + 6 c

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2 a + 5 b - 4 c, a + 4 b - 3 c, 4 a + 7 b - 6 c

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6 a - b - 2 c, 2 a + 3 b + 2 c, - a - 9 b + 7 c

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Solution

The correct options are
A $a - 2 b + 3 c, 2 a + 3 b - 4 c, - 7 b + 10 c$
D $2 a + 5 b - 4 c, a + 4 b - 3 c, 4 a + 7 b - 6 c$
(A) Let P,Q, R be the position vectors.
Let $\to P = \to a - 2 \to b + 3 \to c, \to Q = 2 \to a + 3 \to b - 4 \to c, \to R = - 7 \to b + 10 \to c$
Now $\to P Q = p . v o f Q - p . v o f P$
$= (2 \to a + 3 \to b - 4 \to c) - (\to a - 2 \to b + 3 \to c)$
$= \to a + 5 \to b - 7 \to c$
$\to Q R = p . v o f R - p . v o f Q$
$= (- 7 \to b + 10 \to c) - (2 \to a + 3 \to b - 4 \to c)$
$= - 2 \to a - 10 \to b + 14 \to c$
$= - 2 (\to a + 5 \to b - 7 \to c)$
To test collinearity, $\to P Q = λ \to Q R$
Here, $\to Q R = - 2 \to P Q$
Then, $\to P Q = - \frac{1}{2} \to Q R$
Q is a common point which proves that $P,Q, R are collinear$

(B)Let P,Q, R be the position vectors.
Let $\to P = 3 \to a - 4 \to b + 3 \to c, \to Q = - 4 \to a + 5 \to b - 6 \to c, \to R = 4 \to a - 7 \to b + 6 \to c$
Now $\to P Q = p . v o f Q - p . v o f P$
$= (- 4 \to a + 5 \to b - 6 \to c) - (3 \to a - 4 \to b + 3 \to c)$
$= - 7 \to a + 9 \to b - 9 \to c$
$\to Q R = p . v o f R - p . v o f Q$
$= (4 \to a - 7 \to b + 6 \to c) - (- 4 \to a + 5 \to b - 6 \to c)$
$= 8 \to a - 12 \to b + 12 \to c$
$= 2 (4 \to a - 6 \to b + 6 \to c)$
To test collinearity, $\to P Q = λ \to Q R$
Here, $\to P Q \neq λ \to Q R$
Q is a common point which proves that $P,Q, R are not collinear$

(C) (B)Let P,Q, R be the position vectors.
Let $\to P = 2 \to a + 5 \to b - 4 \to c, \to Q = \to a + 4 \to b - 3 \to c, \to R = 4 \to a + 7 \to b - 6 \to c$
Now $\to P Q = p . v o f Q - p . v o f P$
$= (\to a + 4 \to b - 3 \to c) - (2 \to a + 5 \to b - 4 \to c)$
$= - \to a - \to b + \to c$
$\to Q R = p . v o f R - p . v o f Q$
$= (4 \to a + 7 \to b - 6 \to c) - (\to a + 4 \to b - 3 \to c)$
$= 3 \to a + 3 \to b - 3 \to c$
$= - 3 (- \to a - \to b + \to c)$
To test collinearity, $\to P Q = λ \to Q R$
Here, $\to Q R = - 3 \to P Q$
Q is a common point which proves that $P,Q, R are collinear$

Similarly, if we check for option D, we get $P . Q, R$ are not collinear.

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