Find a terminal point Q of a nonzero vector u = PQ bar with initial point P(-1,

Question

Find a terminal point Q of a nonzero vector u = PQ bar with initial point P(-1, 3, -5) such that a) u is equivalent to the vector v = (6, 7, -3) b) u is oppositely directed to v = (6, 7.-3) (There is more than one answer.) Use the following figure to determine the dot product u middot v assuming the length of the vector u is 3 units and the length of the vector v is 5 units. Consider the planes 2z - y + z = 1 and 1x - 2y + 2z = -2. a) Explain why these planes are parallel. b) Determine the distance between these two parallel planes. Determine a vector x = (a, b, c) which is orthogonal to both u = (1, 0, 1) and = v = (0, 1, 1). [For credit, do not just guess and check: show the method you used to determine x.]

Explanation / Answer

1.Let, the terminal point of u is Q(x,y,z) .

a) Vectors in the same direction must be (positive) scalar multiples of each other. So, we need PQ = tv where t is some scalar ( >0).
There are an infinite number of choices; let's assume t=1.

Thus u = tv

vector PQ = (x+1, y-3, z+5)

u will have the same direction as v .

so, (x+1, y-3, z+5) = (6,7,-3)

=> x = 5, y = 10, z = -8

Thus one possible ans might be Q(5,10,-8)

(b) u will have the opposite direction as v. Vectors in the opposite direction must be negative scalar multiples of each other.There are an infinite number of choices. Let us take t= -1

so, (x+1, y-3, z+5) = (-6,-7,3)

x = -7, y = -4, z = -2

Thus one possible ans might be Q(-7, -4, -2)

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