Let p and q be the position vectors of P and Q respectively with respect to O and -p-p- -q-q- The points R and S divides PQ internally and externally in the ratio 3-2 respectively- If OR and OS are perpendicular- then

Given, OR=3 nbsp; p+2 nbsp; q3+2=15×(3 nbsp; p+2 nbsp; q) OS=3 nbsp; p 2 nbsp; q3 2=(3 nbsp; p 2 nbsp; q) Given that OR and nbsp;OS are perpe ...

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Byju's Answer

Standard XII

Mathematics

Dot Product

Let p and q b...

Question

Let $\to p$ and $\to q$ be the position vectors of $P$ and $Q$ respectively with respect to $O$ and $| \to p | = p, | \to q | = q .$ The points $R$ and $S$ divides $P Q$ internally and externally in the ratio $3 : 2$ respectively. If $- - \to O R$ and $- - \to O S$ are perpendicular, then

9 p^{2} = 4 q^{2}

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4 p^{2} = 9 q^{2}

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9 p = 4 q

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4 p = 9 q

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Solution

The correct option is A $9 p^{2} = 4 q^{2}$
Given,
$- - \to O R = \frac{3 \to p + 2 \to q}{3 + 2} = \frac{1}{5} \times (3 \to p + 2 \to q)$
$- - \to O S = \frac{3 \to p - 2 \to q}{3 - 2} = (3 \to p - 2 \to q)$
Given that $- - \to O R$ and $- - \to O S$
are perpendicular.
$\Rightarrow - - \to O R \cdot - - \to O S = 0$
$\Rightarrow \frac{1}{5} \times (3 \to p + 2 \to q) \cdot (3 \to p - 2 \to q) = 0$
$\Rightarrow 9 | p^{2} | - 4 | q^{2} | = 0$
$\Rightarrow 9 p^{2} = 4 q^{2}$

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Similar questions

Q. Let

\to p

and

\to q

be the position vectors of points

P

and

Q

respectively with respect to origin

(O)

and

| \to p | = p, | \to q | = q .

R, S

divide

P Q

internally and externally in the ratio

2 : 3

respectively. If

- - \to O R

and

- - \to O S

are perpendicular, then

Q. Let

\to p

and

\to q

be the position vectors of

P

and

Q

respectively with respect to

O

and

| \to p | = p, | \to q | = q

R, S

divide

P, Q

Internally and externally in the ratio

2 : 3

respectively. If

O R

and

O S

are perpendicular, then

Q. For any four points

P, Q, R, S,

| - - \to P Q \times - - \to R S - - - \to Q R \times - \to P S + - - \to R P \times - - \to Q S |

is equal to

4

times the area of the triangle

Q. If

\to a, \to b

and

\to c

are three non-coplanar vectors and

\to p, \to q

and

\to r

are vectors defined by

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}

and

\to r = \frac{\to a \times \to b}{[\to a \to b \to c]}

, then the value of

(\to a + \to b) \cdot \to p + (\to b + \to c) \cdot \to q + (\to c + \to a) \cdot \to r =

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\to p, \to q

and

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let

(\to p \times \to q) \times \to r + (\to q \cdot \to r) \to q = (x^{2} + y^{2}) \to q + (14 - 4 x - 6 y) \to p

and

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where

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Let →p and →q be the position vectors of P and Q respectively with respect to O and |→p|=p, |→q|=q. The points R and S divides PQ internally and externally in the ratio 3:2 respectively. If −−→OR and −−→OS are perpendicular, then

The correct option is A 9p2=4q2Given, −−→OR=3→p+2→q3+2=15×(3→p+2→q) −−→OS=3→p−2→q3−2=(3→p−2→q) Given that −−→OR and −−→OS are perpendicular. ⇒−−→OR⋅−−→OS=0 ⇒15×(3→p+2→q)⋅(3→p−2→q)=0 ⇒9|p2|−4|q2|=0 ⇒9p2=4q2

Let $\to p$ and $\to q$ be the position vectors of $P$ and $Q$ respectively with respect to $O$ and $| \to p | = p, | \to q | = q .$ The points $R$ and $S$ divides $P Q$ internally and externally in the ratio $3 : 2$ respectively. If $- - \to O R$ and $- - \to O S$ are perpendicular, then