Find the component form of v vector given its magnitude and the angle it makes w

Question

Find the component form of v vector given its magnitude and the angle it makes with the positive x-axis. ||v^rightarrow|| = 2, theta = 150 degree Complete the square to write the equation of the sphere in standard form. Find the center and radius. x^2 + y^2 + z^2 - 2x + 6y + 8z + 1 = 0 Find 2 u vector + 5v vector for u vector = lang 4, 9 rang, v vector = lang 2, -5 rang Find a unit vector in the direction of v vector v vector = lang 2, -1, 2 rang Find the component form of the vector v vector with the initial point: (-1, 2, 3) and terminal point: (3, 3, 4) Determine whether u vector = lang 2, -3, 1 rang and v vector = lang 1, -1, -1 rang are orthogonal, parallel or neither. Find the direction cosines of u vector = 3i vector + 2j vector - 2k vector and demonstrate that the sum of the squares of the direction cosines is 1. Given u vector = 2i vector + 3j vector and v vector = 5i vector + 7j vector (a) Find the projection of u vector onto v vector (b) Find the vector component of u vector orthogonal to v vector Find the area of the parallelogram that has the vectors u vector = j^rightarrow, and v vector = j vector + k vector as adjacent sides. Find u vector middot (v vector times w^rightarrow) u vector = lang 2, 0, 1 rang v vector = lang 0, 3, 0 rang w vector = lang 0, 0, 1 rang

Explanation / Answer

First off, we are to solve only one question per problem that u ask.

But ive gone ahead and solved four of them.

Here they are.

1)
|v| = 2
theta = 150

Horizontal component : |v|cos(theta)
= 2 * cos(150)
= 2 * -sqrt(3)/2
= -sqrt3

Vertical component :
|v|sintheta
2 * sin(150)
2 * 1/2
= 1

So, the vector is :

-sqrt(3) i + 1j ----> ASNWER

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2)

x^2 - 2x + y^2 + 6y + z^2 + 8z = -1

x^2 - 2x + 1 + y^2 + 6y + 9 + z^2 + 8z + 16 = -1 + 1 + 9 + 16

(x - 1)^2 + (y + 3)^2 + (z + 4)^2 = 25

So, center = (1 , -3 , -4)
Radius = sqrt25 = 5

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3)
2u = 2<4,9> = <8,18>
5v = 5<2,-5> = <10,-25>

2u + 5v becomes :

<8,18> + <10,-25>

Adding them gives :
<18 , -7>

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4)

v = <2 , -1 , 2>

|v| = sqrt(a^2 + b^2 + c^2)

So, |v| = magnitude of vector v= sqrt(2^2 + (-1)^2 + 2^2)

|v| = sqrt(4 + 1 + 4)

|v| = sqrt9

|v| = 3

Sos, unit vector in the direction of a vector v is v / |v|

So, we have :

<2 , -1 , 2> / 3

So, the required unit vector is

<2/3 , -1/3 , 2/3>

Hope this made sense!

Cheers!

Plz ask the remaining as a separate question.

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