A vector $\overrightarrow a$ has components 3p and 1 with respect to a 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, $\overrightarrow a$ has components p + 1 and $\sqrt {10}$, then the value of p is equal to :
Solution
$${\left| {\overrightarrow a } \right|_{old}} = {\left| {\overrightarrow a } \right|_{new}}$$<br><br>(3p)<sup>2</sup> + 1 = (p + 1)<sup>2</sup> + 10<br><br>$\Rightarrow$ 9p<sup>2</sup> $-$ p<sup>2</sup> $-$ 2p $-$ 10 = 0<br><br>$\Rightarrow$ 8p<sup>2</sup> $-$ 2p $-$ 10 = 0<br><br>$\Rightarrow$ 4p<sup>2</sup> $-$ p $-$ 5 = 0<br><br>$\Rightarrow$ 4p<sup>2</sup> $-$ 5p + 4p $-$ 5 = 0<br><br>$\Rightarrow$ (4p $-$ 5) (p + 1) = 0<br><br>$\Rightarrow$ p = ${5 \over 4}$, $-$ 1
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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