Vector product operation of vector? Note: the following letters are all vectors, except K, where x is cross product M = 2A + B N = Ka + B │ a │ = 1 │ B │ = 2 a ⊥ B Q: when k is the value, the area of parallelogram with m and N as adjacent sides is 6 │ m × n │ = (2a + b) × (KA + b) │ = (2 + k) * a × B │ = │ 2 + K │ a × B │ = 2 │ 2 + K │ = 6 to get k = 1 or 5, but the answer after class is - 1 or 5. What's the matter?

Vector product operation of vector? Note: the following letters are all vectors, except K, where x is cross product M = 2A + B N = Ka + B │ a │ = 1 │ B │ = 2 a ⊥ B Q: when k is the value, the area of parallelogram with m and N as adjacent sides is 6 │ m × n │ = (2a + b) × (KA + b) │ = (2 + k) * a × B │ = │ 2 + K │ a × B │ = 2 │ 2 + K │ = 6 to get k = 1 or 5, but the answer after class is - 1 or 5. What's the matter?

If your book is yellow, there is a formula in the middle of page 16 of your book: a × B = - B × A. then this question: m × n = (2a + b) × (KA + b) │ = │ 2K * a × a + 2A × B + B × k * a + B × B │ = │ 0 + 2A × B-K * a × B + 0 │ = (2-k) (a × b) │ You can do it later!

Given vector a = (√ 2 √ 2), try to find the unit vector with 45 ° angle with vector a

There are two unit vectors with an angle of π / 4 with a = (√ 2, √ 2)
(1,0) and (0,1)
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Let B = (x, y), then: x ^ 2 + y2 = 1
b·a=(x,y)·(√2,√2)=√2x+√2y
So: cos = a · B / (| a | * | B |) = (√ 2x + √ 2Y) / (2sqrt (x ^ 2 + y ^ 2)) = √ 2 / 2
Namely: sqrt (x ^ 2 + y ^ 2) = x + y
That is: xy = 0, that is: x = 0 or y = 0
That is: B = (1,0) or (0,1)

Given a vector = (x1, y1-1), B vector = (X2, y2-2), then a vector * B vector is equal to?

a·b=(x1,y1-1)·(x2,y2-2)
=x1x2+(y1-1)(y2-2)
=x1x2+y1y2+2-2y1-y2

Vector a = (x1, Y1). Vector b = (X2, Y2). What is the absolute value of vector A-B equal to

|Vector a-vector B|
=|(x1,y1)-(x2,y2)|
=|(x1-x2,y1-y2)|
=√[(x1-x2)^2+(y1-y2)^2]