A vector veca has components 3 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, veca has components p+1 and √10, then a value of p is equal to:

Question

A vector veca has components 3 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, veca has components p+1 and √10, then a value of p is equal to: (A) 1 (B) -(5/4) (C) (4/5) (D) -1

Tardigrade Admin · Accepted Answer

a text Old =3 p hati+ hatj a text New =(p+1) hati+√10 hatj ⇒|aOld|=|a text New | <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/a9f9c13987f50a23053a47a5d0d14f85-.png" /> ⇒ ap2+1=p2+2 p+1+10 8 p2-2 p-10=0 4 p2-p-5=0 (4 p -5)( p +1)=0 arrow p =(5/4),-1

Q. A vector $a$ has components 3 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, $a$ has components $p + 1$ and $10$ , then a value of p is equal to:

1

$- \frac{5}{4}$

$\frac{4}{5}$

-1

$a_{Old} = 3 p \hat{i} + \hat{j}$
$a_{New} = (p + 1) \hat{i} + 10 \hat{j}$
$\Rightarrow ∣ ∣ a_{Ol d} ∣ ∣ = ∣ ∣ a_{New} ∣ ∣$

$\Rightarrow a p^{2} + 1 = p^{2} + 2 p + 1 + 10$
$8 p^{2} - 2 p - 10 = 0$
$4 p^{2} - p - 5 = 0$
$(4 p - 5) (p + 1) = 0 \to p = \frac{5}{4}, - 1$

Q. A vector a has components 3 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, a has components p+1 and 10​, then a value of p is equal to:

1

−45​

54​

-1

aOld ​=3pi^+j^​ aNew ​=(p+1)i^+10​j^​ ⇒∣∣​aOld​∣∣​=∣∣​aNew ​∣∣​ ⇒ap2+1=p2+2p+1+10 8p2−2p−10=0 4p2−p−5=0 (4p−5)(p+1)=0→p=45​,−1

Q. A vector $a$ has components 3 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, $a$ has components $p + 1$ and $10$ , then a value of p is equal to:

$- \frac{5}{4}$

$\frac{4}{5}$

$a_{Old} = 3 p \hat{i} + \hat{j}$
$a_{New} = (p + 1) \hat{i} + 10 \hat{j}$
$\Rightarrow ∣ ∣ a_{Ol d} ∣ ∣ = ∣ ∣ a_{New} ∣ ∣$

$\Rightarrow a p^{2} + 1 = p^{2} + 2 p + 1 + 10$
$8 p^{2} - 2 p - 10 = 0$
$4 p^{2} - p - 5 = 0$
$(4 p - 5) (p + 1) = 0 \to p = \frac{5}{4}, - 1$