Solve for j:
6-8 i j+(10 j^2)/3 = 0
The left hand side factors into a product with four terms:
2/3 (-3 i+j) (3 i+5 j) = 0
Multiply both sides by 3/2:
(-3 i+j) (3 i+5 j) = 0
Split into two equations:
-3 i+j = 0 or 3 i+5 j = 0
Subtract -3 i from both sides:
j = 3 i or 3 i+5 j = 0
Subtract 3 i from both sides:
j = 3 i or 5 j = -3 i
Divide both sides by 5:
Answer: | j = 3 i or j = -(3 i)/5
I might help with your problem if you help with the problem that says home please help and explain
+495 v=√34 cis59.04°
w = 2√10/3 cis -18.43°
v*w = 2√10/3 * √34 cis59.04°-18.43°
= 12.293(cos40.61°)+i(sin40.61)
= 9.332i + 8.002j
*answer may be subject to rounding errors*
I might put all that in radical form later.
Solve for j:
6-8 i j+(10 j^2)/3 = 0
The left hand side factors into a product with four terms:
2/3 (-3 i+j) (3 i+5 j) = 0
Multiply both sides by 3/2:
(-3 i+j) (3 i+5 j) = 0
Split into two equations:
-3 i+j = 0 or 3 i+5 j = 0
Subtract -3 i from both sides:
j = 3 i or 3 i+5 j = 0
Subtract 3 i from both sides:
j = 3 i or 5 j = -3 i
Divide both sides by 5:
Answer: | j = 3 i or j = -(3 i)/5
+33661 I assume i and j are unit vectors in the x- and y-directions respectively and that a dot product is required for v.w.
If so, then i.i = i, j.j = j and i.j = j.i = 0 so:
(3i + 5j)(-2i + 2/3j) = -6i.i + 2i.j - 10j.i +10/3j.j = -6i + 10/3j
If v*w is meant to be a cross product (vxw) then the third, z, dimension is involved and
ixi = jxj = kxk = 0
ixj = -jxi = k
jxk = -kxj = i
kxi = -ixk = j
