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Vector Algebra Part 1

Section: Mathematics

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Question : 6 of 100

Marks: +1, -0

A vector a has components

3 p

and 1 with respect to rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, a has components

p+1

\sqrt{10}

, then a value of

p

[18 Mar 2021 Shift 1]

$1$
$-\;\frac{5}{4}$
$\;\frac{4}{5}$
$-1$

Solution:

After counter clockwise (or anti-clockwise) rotation, the length of the vector a remains constant.

i.e.

|a|

at old position

=|a|

at new position

\Rightarrow \;\; (3p)^{2}+(1)^{2}=(p+1)^{2}+(\sqrt{10})^{2}

\Rightarrow \;\; 9p^{2}+1=p^{2}+1+2p+10

\Rightarrow \;\; 8p^{2}-2p-10=0 \Rightarrow 4p^{2}-p-5=0

\Rightarrow \;\; (p+1)(4p-5)=0

\Rightarrow \;\; p=\frac{5}{4},-1

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