JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 7)
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 :
1
$$ - {5 \over 4}$$
$${4 \over 5}$$
$$-$$1
Explanation
$${\left| {\overrightarrow a } \right|_{old}} = {\left| {\overrightarrow a } \right|_{new}}$$
(3p)2 + 1 = (p + 1)2 + 10
$$ \Rightarrow $$ 9p2 $$-$$ p2 $$-$$ 2p $$-$$ 10 = 0
$$ \Rightarrow $$ 8p2 $$-$$ 2p $$-$$ 10 = 0
$$ \Rightarrow $$ 4p2 $$-$$ p $$-$$ 5 = 0
$$ \Rightarrow $$ 4p2 $$-$$ 5p + 4p $$-$$ 5 = 0
$$ \Rightarrow $$ (4p $$-$$ 5) (p + 1) = 0
$$ \Rightarrow $$ p = $${5 \over 4}$$, $$-$$ 1
(3p)2 + 1 = (p + 1)2 + 10
$$ \Rightarrow $$ 9p2 $$-$$ p2 $$-$$ 2p $$-$$ 10 = 0
$$ \Rightarrow $$ 8p2 $$-$$ 2p $$-$$ 10 = 0
$$ \Rightarrow $$ 4p2 $$-$$ p $$-$$ 5 = 0
$$ \Rightarrow $$ 4p2 $$-$$ 5p + 4p $$-$$ 5 = 0
$$ \Rightarrow $$ (4p $$-$$ 5) (p + 1) = 0
$$ \Rightarrow $$ p = $${5 \over 4}$$, $$-$$ 1
Comments (0)
