The position vector of P is 4-3- and the position vector of Q is -2-5- The vector 2PQ is

Since, the position vector is taken from origin O. So, we have PQ=OQ OP ⇒ PQ=( 2î+5ĵ) (4î+3ĵ)= 6î+2ĵ Thus, 2PQ= 2( 6î+2ĵ)= 12î+4ĵ ...

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Standard XI

Physics

Introduction

The position ...

Question

The position vector of $P$ is $4^i + 3^j$ and the position vector of $Q$ is $- 2^i + 5^j$ . The vector $2 - - \to P Q$ is

- 6^i + 2^j

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- 12^i + 4^j

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12^i - 4^j

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6^i - 2^j

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Solution

The correct option is B $- 12^i + 4^j$
Since, the position vector is taken from origin $O$ . So, we have
$- - \to P Q = - - \to O Q - - - \to O P$
$\Rightarrow - - \to P Q = (- 2^i + 5^j) - (4^i + 3^j) = - 6^i + 2^j$
Thus, $2 - - \to P Q = 2 (- 6^i + 2^j) = - 12^i + 4^j$

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Similar questions

Q. The position vector of

P

4^i + 3^j

and the position vector of

Q

- 2^i + 5^j

. The vector

2 - - \to P Q

Q. Forces

2^i + 7^j, 2^i - 5^j + 6^k, -^i + 2^j -^k

act at a point

P

whose position vector is

4^i - 3^j - 2^k

. Find the vector moment of the resultant of three forces acting at

P

about the point

Q

, whose position vector is

6^i +^j - 3^k

Q. Let

¯ ¯ ¯ a = 4^i + 7^j + 8^k, ¯ ¯ b = 2^i + 3^j + 4^k a n d ¯ ¯ c = 2^i + 5^j + 7^k

are the position vectors of triangle

A B C .

The position vectors of the point where the bisector of angle

A

meets at

B C

, is

Q. If the position vectors of the vertices

A, B

and

C

of a

△ A B C

are respectively

4^i + 7^j + 8^k, 2^i + 3^j + 4^k

and

2^i + 5^j + 7^k

, then the position vector of the point, where the bisector of

∠ A

meets

B C

is :

Q. If

4^i + 7^j + 8^k, 2^i + 3^j + 4^k

and

2^i + 5^j + 7^k

are the position vectors of the vertices

A, B

and

C

, respectively of triangle

A B C

, then the position vector of the point where the bisector of

∠ A

meets

B C

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The position vector of P is 4^i+3^j and the position vector of Q is −2^i+5^j. The vector 2−−→PQ is

The correct option is B −12^i+4^jSince, the position vector is taken from origin O. So, we have −−→PQ=−−→OQ−−−→OP ⇒ −−→PQ=(−2^i+5^j)−(4^i+3^j)=−6^i+2^j Thus, 2−−→PQ=2(−6^i+2^j)=−12^i+4^j

The position vector of $P$ is $4^i + 3^j$ and the position vector of $Q$ is $- 2^i + 5^j$ . The vector $2 - - \to P Q$ is

The correct option is B $- 12^i + 4^j$
Since, the position vector is taken from origin $O$ . So, we have
$- - \to P Q = - - \to O Q - - - \to O P$
$\Rightarrow - - \to P Q = (- 2^i + 5^j) - (4^i + 3^j) = - 6^i + 2^j$
Thus, $2 - - \to P Q = 2 (- 6^i + 2^j) = - 12^i + 4^j$