Question

Find the value of a+b , if the points (2,a,3), (3,-5,b), (-1,11,9) are collinear

Answer 1

a + b = 0

Step-by-step explanation:

Given:

Let the three given position vectors be $A(2\hat{i}+a\hat{j}+3\hat{k}), B(3\hat{i}-5\hat{j}+b\hat{k}), C(-\hat{i}+11\hat{j}+9\hat{k})$

$\vec{AB}= (3-2)\hat{i}+(-5-a)\hat{j}+(b-3)\hat{k}$

$\vec{AB}= \hat{i}+(-5-a)\hat{j}+(b-3)\hat{k}$

$\vec{BC}= (-1-3)\hat{i}+(11+5)\hat{j}+(9-b)\hat{k}$

$\vec{BC}= (-4)\hat{i}+(16)\hat{j}+(9-b)\hat{k}$

∵ A,B and C are collinear vectors,

$\frac{1}{-4} =\frac{-5-a}{16} =\frac{b-3}{9-b}$

⇒ $\frac{-5-a}{16}=-\frac{1}{4}$    

⇒ 4 (-5-a) = -16    

⇒ -20 - 4a = -16

⇒ -4a = -16 +20

⇒ -4a = 4

⇒ a= -1

And  

⇒ $\frac{b-3}{9-b} =-\frac{1}{4}$

⇒ 4(b-3) = - (9-b)

⇒ 4b - 12 = -9 + b

⇒ 4b - b = -9 +12

⇒ 3b = 3

⇒ b = 1

Now,

a + b = -1 + 1 = 0