Question

Solve the Question in Attachment

Options:

a) t = 2

b) t = 4

c) t = 6

d) t = 8

Thank You​

Answer 1

Answer:

Given :

Two vectors A and B .

To find :

The value of "t".

Concept:

When 2 vectors are orthogonal, then there dot product is zero (0).

Formulas used:

sin(a - b) = sin(a)cos(b) - sin(b)cos(a)

Calculation:

Angle between the vectors = 90°

Dot product/Scalar product of the 2 given vectors:

A.B = |A| × |B| × cos(90°) = 0

=> sin(αt)cos(αt²/4) - cos(α)sin(αt²/4) = 0

=> sin(αt - αt²/4) = 0

=> αt - αt²/4 = 0

Cancelling "α" on both sides :

=> t - t²/4 = 0

=> t(1- t/4) = 0

so "t" can be either 0 or

1 - t/4 = 0

=> t/4 = 1

=> t = 4.

So the answer is t = 0 or 4.

OPTION b) IS THE CORRECT ANSWER

Answer 2

ANSWER :- t=4

SOLUTION: -

GIVEN: - The two vectors are orthogonal to each other.

REQUIRED TO FIND :- Value of t.

CONCEPT :-

The dot product of two orthogonal vectors is zero.

FORMULAS :-

☆ Sin(a-b) = sin(a)cos(b)-sin(b)cos(a).-----(1)

☆ Dot product of 2 vectors => A.B = |A||B|cosX------(2)

STEPS :-

The angle between two vectors is 90.

From 2,

=> |A||B|cos90 =0.

From 1,

$= > \sin( \alpha t) \cot( \alpha {t}^{2} \div 4 ) - \cos( \alpha t) \sin( \alpha t {}^{2} \div 4 )$

$= > \sin( \alpha t - \alpha t {}^{2} \div 4 ) = 0$

$= > \alpha t - \alpha t {}^{2} \div 4 = 0$

$= > \alpha t = \alpha t {}^{2} \div 4$

$= > t \div 4 = 1$

$= > t = 4$

Hence, t =4.

HOPE IT HELPS :)