Let vector a = 3i + j - k.

Find the vector b in the direction of vector a such that |b| = 5

5i +(5/3)j - (5/3)k

Step-by-step explanation:

Vector b is in the same direction as vector a, so it is a scalar multiple of a.

b = ca for some positive real number c.

b = c(3i + j - k) = 3ci + cj - ck

|b| = sqrt[(3c)^2 + c^2 - c^2] = sqrt(9c^2) = 3c

If |b| = 5, then 3c = 5 and c = 5/3

That makes b = (5/3)(3i + j - k) = 5i +(5/3)j - (5/3)k

i you goofed up when writing this, because hnot is not a word minusf is not either. i cannot answer   because of that.

step-by-step explanation:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3