Question

If the vector v = (-2,1,2) is rotated -45° along y-axis, then image of v is
a) (-√2,1,√2)
b)(-√2,0,1)
c) (-2√2,1,0)
d) (0,1,-2√2)

Answer 1

Given :

Vector v = ( -2,1,2 )

Angle = -45° along y axis.

To find :

Image of v.

Solution :

For rotating a vector along an axis, a rotation matrix is used, to perform the rotation of that vector.

Rotation along different axis have different formulas :

Let, new vector is (X,Y,Z)

To find the image of a vector along y direction,

we have a formula :

$X = x \cos( \theta) + z \sin( \theta) \\ Y = y \\ Z = - x \sin( \theta) + z \cos( \theta)$

(equation 1)

From given data we know that

x = -2 ,

y = 1 ,

z = 2 ,

θ = -45°

Putting all these values in equation 1.

$X = - 2\cos(-45° ) + 2 \sin(-45° ) \\ Y = 1 \\ Z = - ( - 2) \sin(-45° ) + 2 \cos(-45° ) \:$

we know the values of these trigonometric functions as :

$\cos(-45°) = \frac{1}{ \sqrt{2} } \\ \sin(-45°) = - \frac{1}{ \sqrt{2} }$

so,

Putting again in above equation, we get the values of X,Y and Z as :

$X = - 2( \frac{1}{ \sqrt{2} }) + 2 ( \frac{ - 1}{ \sqrt{2} } ) \\ Y = 1 \\ Z = - ( - 2) ( \frac{ - 1}{ \sqrt{2} } ) + 2 ( \frac{1}{ \sqrt{2} }) \:$

On solving this, we get

$\bf \: X = - \frac{4}{ \sqrt{2} } = - 2 \sqrt{2} \\ \bf Y = 1 \\ \bf Z = - \frac{2}{ \sqrt{2} } + \frac{2}{ \sqrt{2} } = 0$

So image of given vector is :

(X,Y,Z) = (-2√2 , 1 , 0 )

Answer 2

Answer:

Step-by-step explanation:

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