Question

the simple sum of two coinitial vectors is 16 units.their vectors sum is 8 units . The resultant of the vectors is perpendicular to the smaller vector.the magnitudes of the two vectors are​

Answer 1

Answer:6 Units and 10 Units

Explanation:

By Resultant Formula

SEE BELOW THE Attachment and like this answer

Answer 2

The magnitudes of the two vectors are 10 units and 6 units respectively.

The simple sum of two co - initial vectors is 16 units. their vector sum is 8 units. The resultant of the vectors is perpendicular to the smaller vector.

The magnitudes of the two vectors are 10 units and 6 units respectively.

We have to find the magnitudes of the two vectors.

Let P and Q are two vectors such that P is greater than Q and θ is the angle between them.

A/C to question,

Simple of two vectors = 16

⇒ P + Q = 16 ....(1)

Vector sum = R = 8

Also dot product of resultant and smaller vector Q equals zero [∵ resultant is perpendicular to the smaller vector ]

⇒ R.Q = 0

⇒ (P + Q).Q = 0

⇒ P.Q + Q.Q = 0

⇒ P.Q + Q² = 0

⇒ P.Q = - Q² ...(2)

Now resultant vector of P and Q is given by,

$\implies\sqrt{P^2+Q^2+2P.Q}=8\\\\\implies P^2+Q^2+2(-Q^2)=8^2\\\\\implies P^2-Q^2=8^2\\\\\implies(P-Q)(P+Q)=64\\\\\implies(P-Q)16=64\\\\\implies(P-Q)=4\:\:....(3)$

From equation (1) and (3) we get,

P = 10 and Q = 6

Therefore the magnitudes of the two vectors are 10 units and 6 units respectively.

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