What is {{ c}} equal to?

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

{{3≤ft( {{{ i}} + {{ j}}} )}}{2}

{{2≤ft( {{{ i}} + {{ j}}} )}}{3}

{{≤ft( {{{ i}} + {{ j}}} )}}{2}

{{≤ft( {{{ i}} + {{ j}}} )}}{3}

What is {{ c}} equal to?

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

Test Index

UPSC NDA 2 2016 Mathematics Solved Paper

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Question Numbers: 65-66

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

Let

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

and

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

where

{\rm{\vec c}}

is parallel to

{\rm{\vec a}}

and

{\rm{\vec d}}

is perpendicular to

{\rm{\vec a}}

Question : 65 of 120

Marks: +1, -0

What is

{\rm{\vec c}}

equal to?

$\frac{{3\left( {{\rm{\hat i}} + {\rm{\hat j}}} \right)}}{2}$
$\frac{{2\left( {{\rm{\hat i}} + {\rm{\hat j}}} \right)}}{3}$
$\frac{{\left( {{\rm{\hat i}} + {\rm{\hat j}}} \right)}}{2}$
$\frac{{\left( {{\rm{\hat i}} + {\rm{\hat j}}} \right)}}{3}$

Validate

Solution:

Concept:

Parallel vectors satisfy

\vec c = \lambda \vec a

and perpendicular vectors satisfy

\vec d \cdot \vec a = 0

Explanation:

Given

\vec a = \hat i + \hat j

\vec b = 3\hat i + 4\hat k

, and

\vec b = \vec c + \vec d

Since

\vec c

is parallel to

\vec a

, let

\vec c = \lambda \vec a

Since

\vec d

is perpendicular to

\vec a

, we have

\vec d \cdot \vec a = 0

Take dot product of both sides of

\vec b = \vec c + \vec d

with

\vec a

\vec b \cdot \vec a = (\vec c + \vec d) \cdot \vec a = \vec c \cdot \vec a + \vec d \cdot \vec a = \lambda (\vec a \cdot \vec a) + 0 = \lambda |\vec a|^{2}

Compute

\vec b \cdot \vec a = (3\hat i + 4\hat k) \cdot (\hat i + \hat j) = 3(1) + 0 + 0 = 3

Compute

|\vec a|^{2} = 1^{2} + 1^{2} = 2

Thus

3 = \lambda (2)

gives

\lambda = \frac{3}{2}

Therefore

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

Answer:

\frac{3\left( \hat i + \hat j \right)}{2}

(Option A).

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