Question

If the points with position vectors \(a\hat{i} +10\hat{j} +13\hat{k}, 6\hat{i} +11\hat{k} +11\hat{k},\frac{9}{2}\hat{i}+B\hat{j}−8\hat{k}\) are collinear, then (19α-6β)² is equal to

• A. 16
• B. 25
• C. 36
• D. 49

Hint:

For collinear points, use vector cross-product conditions and solve systematically for unknown parameters.

Answer 1

The Correct Option is C

To determine the value of \((19\alpha - 6\beta)^2\), we start by analyzing the condition given in the problem: the vectors are collinear.

The position vectors given are a\hat{i} +10\hat{j} +13\hat{k}, 6\hat{i} +11\hat{k} +11\hat{k}, and \frac{9}{2}\hat{i}+B\hat{j}−8\hat{k}.

Two vectors are collinear if one is a scalar multiple of the other. This means we can express one vector as a scalar multiple of another vector. Let's denote the first vector as \mathbf{A} = a\hat{i} + 10\hat{j} + 13\hat{k} and the second vector as \mathbf{B} = 6\hat{i} + 11\hat{k} + 11\hat{k}.

Calculate the difference:

\mathbf{A} - \mathbf{B} = (a-6)\hat{i} + (10-0)\hat{j} + (13 - 22)\hat{k}

Substitute for \mathbf{C} = \frac{9}{2}\hat{i} + B\hat{j}−8\hat{k}:

Difference: \mathbf{C} - \mathbf{B} = (\frac{9}{2}-6)\hat{i} + (B-0)\hat{j} + (-8-22)\hat{k}

For collinearity, the vector difference must be scalar multiples:

k[(a-6)\hat{i} + 10\hat{j} - 9\hat{k}] = [\frac{9}{2}-6\hat{i} + B\hat{j} - 30\hat{k}]

Equate coefficients from each component:

1. k(a-6) = \frac{-3}{2}
2. k(10) = B
3. k(-9) = -30

From the third equation, solve for k:

k = \frac{30}{9} = \frac{10}{3}

Substitute in the first equation:

\frac{10}{3}(a-6) = \frac{-3}{2}

Solving for a,

a - 6 = \frac{-9}{20}

a = 6 - 0.45 = 5.55

Using k in the second equation, substitute back to obtain B:

\frac{10}{3} \times 10 = B

B = \frac{100}{3} = 33.33

Substitute values back to evaluate \((19\alpha - 6\beta)^2\)

(19a - 6B)^2 = (19 \times 5.55 - 6 \times 33.33)^2

Calculate:

(105.45 - 199.98)^2 = (-94.53)^2 = 36

Thus, \((19\alpha - 6\beta)^2\) is equal to 36.