{ 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

NDA Math Vector Algebra

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

Let

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

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 : 61 of 101

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}) + \hat{xi} + \hat{yj} + \hat{zk}

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

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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