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Vector Algebra Part 1

Section: Mathematics

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Question : 12 of 100

Marks: +1, -0

Let

a=\hat{i}+2\hat{j}-3\hat{k}

b=2\hat{i}-3\hat{j}+5\hat{k}

r \times a = b \times r, r \cdot (\alpha \hat{i} + 2 \hat{j} + \hat{k}) = 3

(2 \hat{i} + 5 \hat{j} - a \hat{k}) = -1, \alpha \in R

, then the value of

\alpha + |r|^2

[16 Mar 2021 Shift 2]

9
15
13
11

Solution:

Given,

a=(\hat{i}+2\hat{j}-3\hat{k})

and

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

If

r \times a = b \times r, r \cdot (\alpha \hat{i} + 2 \hat{j} + \hat{k}) = 3

r \cdot (2 \hat{i} + 5 \hat{j} - \alpha \hat{k}) = -1

\Rightarrow r \times a = b \times r

(r \times a) = - (r \times b) \Rightarrow (r \times a) + (r \times b) = 0

\Rightarrow r \times (a + b) = 0 \Rightarrow r = \lambda (a + b)

\Rightarrow \;\; r = \lambda [(1+2) \hat{i} + (2-3) \hat{j} + (-3+5) \hat{k}]

\Rightarrow r = \lambda (3 \hat{i} - \hat{j} + 2 \hat{k})

r \cdot (\alpha \hat{i} + 2 \hat{j} + \hat{k}) = 3

\Rightarrow \lambda (3 \hat{i} - \hat{j} + 2 \hat{k}) (\alpha \hat{i} + 2 \hat{j} + \hat{k}) = 3

\Rightarrow \;\; \lambda (3\alpha - 2 + 2) = 3

\Rightarrow \alpha \lambda = 1

r \cdot (2 \hat{i} + 5 \hat{j} - \alpha \hat{k}) = -1

\Rightarrow \lambda (3 \hat{i} - \hat{j} + 2 \hat{k}) (2 \hat{i} + 5 \hat{j} - \alpha \hat{k}) = -1

\Rightarrow \;\; \lambda (6 - 5 - 2\alpha) = -1 \Rightarrow \lambda (1 - 2\alpha) = -1

\Rightarrow \;\; \lambda - 2\alpha\lambda = -1 \Rightarrow \lambda - 2 = -1

\Rightarrow \;\; \lambda = 1

So,

\alpha = 1

r = (3 \hat{i} - \hat{j} + 2 \hat{k})

\Rightarrow \;\; |r|^2 = 9 + 1 + 4 = 14

\therefore \;\; \alpha + |r|^2 = 1 + 14 = 15

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