Solve this following

Question:

A vector $\vec{a}$ has components $3 \mathrm{p}$ 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, $\vec{a}$ has components $p+1$

and $\sqrt{10}$, then a value of $p$ is equal to:

  1. 1

  2. $-\frac{5}{4}$

  3. $\frac{4}{5}$

  4. $-1$


Correct Option: , 4

Solution:

$\vec{a}_{\text {Old }}=3 p \hat{i}+\hat{j}$

$\vec{a}_{\text {New }}=(p+1) \hat{i}+\sqrt{10} \hat{j}$

$\Rightarrow\left|\vec{a}_{\text {Old }}\right|=\left|\vec{a}_{\text {New }}\right|$

$\Rightarrow a p^{2}+1=p^{2}+2 p+1+10$

$8 \mathrm{p}^{2}-2 \mathrm{p}-10=0$

$4 \mathrm{p}^{2}-\mathrm{p}-5=0$

$(4 \mathrm{p}-5)(\mathrm{p}+1)=0 \rightarrow \mathrm{p}=\frac{5}{4},-1$

 

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