\; { What is }\; {c}\; { equal to? }\; [NDA II 2016]

Correct Answer is : 3(i^+j^)2{3({i}+{j})}{2}23(i^+j^)

Correct Answer is : 3(i^+j^)2{3({i}+{j})}{2}23(i^+j^)

Test Index

UPSC NDA 2 2016 Mathematics Solved Paper

Consider the following for the next two (02) items that follow:

Let

\vec{a} = \hat{i} + \hat{j}, \vec{b} = 3\hat{i} + 4\hat{k}

and

\vec{b} = \vec{c} + \vec{d}

,where

\vec{c}

is parallel to

\vec{a}

and

\vec{d}

is perpendicular to

\vec{a}

[NDA II 2016]

Question : 65 of 120

Marks: +1, -0

\; \text{ What is }\; \vec{c}\; \text{ equal to? }\;

Solution:

Since

\vec{c}

is parallel to

\vec{a}

\vec{c}=\lambda\vec{a}

Now

\vec{b}=\vec{c}+\vec{d}=\lambda\vec{a}+\vec{d}

=\lambda(\hat{i}+\hat{j})+x\hat{i}+y\hat{j}+z\hat{k}

3\hat{i}+4\hat{k}=(\lambda+x)\hat{i}+(\lambda+y)\hat{j}+z\hat{k}

Comparing we get

z=4,\lambda+y=0,\lambda+x=3\Rightarrow -y+x=3

(From(1))

\Rightarrow\lambda=-y

...(1)

\Rightarrow x-y=3

....(2)

Now

\vec{d}

\perp r

\vec{a}

So,

\cos\theta=0

\Rightarrow x+y=0

.....(3)

Solving (2) and (3) we get

2x=3

\Rightarrow x=\frac{3}{2}, y=-\frac{3}{2}

\Rightarrow\vec{c}=\lambda(\vec{a})=\frac{3}{2}(\hat{i}+\hat{j})

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