Video: Finding the Unknown Component of a Vector given the Result of Its Cross Product by Another Vector

If 𝐀 = 3𝐒 βˆ’ 5𝐣, 𝐁 = π‘šπ’ + 5𝐣, and 𝐀 Γ— 𝐁 = 50𝐀, find the value of π‘š.

02:24

Video Transcript

If the vector 𝐀 is equal to three 𝐒 minus five 𝐣, 𝐁 is equal to π‘šπ’ plus five 𝐣, and the cross product of 𝐀 and 𝐁 is equal to 50𝐀, find the value of π‘š.

To answer this question, we’re going to need to recall what we mean by the cross product of two vectors. Essentially, it’s a way of multiplying two vectors. Unlike the dot product, where we get a scalar quantity, when we find the cross product of two vectors, we get a vector. And we say that if 𝐚 is the three-dimensional vector given by π‘Ž one, π‘Ž two, π‘Ž three and 𝐛 is the vector 𝑏 one, 𝑏 two, 𝑏 three. Then the cross product of 𝐚 and 𝐛 is equal to the vector π‘Ž two 𝑏 three minus π‘Ž three 𝑏 two, π‘Ž three 𝑏 one minus π‘Ž one 𝑏 three, and π‘Ž one 𝑏 two minus π‘Ž two 𝑏 one.

Now, this isn’t a particularly easy formula to remember. And if you’re struggling, you might want to recall that it’s really just the determinant of a three-by-three matrix. So now, we have a formula. Let’s define π‘Ž one, π‘Ž two, π‘Ž three and 𝑏 one, 𝑏 two, 𝑏 three. π‘Ž one is the horizontal component for 𝐚. It’s three. π‘Ž two is the vertical component. That’s negative five. And there’s no 𝐀-component. So π‘Ž three is equal to zero. 𝑏 one is equal to π‘š. 𝑏 two is equal to five. And once again, 𝑏 three is equal to zero.

The first element of the cross product of these two vectors then is π‘Ž two 𝑏 three. So that’s negative five times zero minus π‘Ž three 𝑏 two, which is zero multiplied by five. The second element is π‘Ž three 𝑏 one, which is zero times π‘š, minus π‘Ž one 𝑏 three, which is three times zero. And the third element is π‘Ž one 𝑏 two, that’s three times five, minus π‘Ž two 𝑏 one, that’s negative five multiplied by π‘š. The first and second elements simplify to zero. And the third element of the cross product of our two vectors becomes 15 plus five π‘š. And that’s because we have minus negative five. So we get plus five.

Now, if we go back to the question, we see that the cross product of 𝐀 and 𝐁 is given to us. We’re told it’s 50𝐀. Well, another way to represent that is using these sort of triangular brackets. The vector 𝐚 crossed with 𝐛 is zero, zero, 50. And we see then that, for the vectors to be equal, 50 must be equal to 15 plus five π‘š. We’ll solve by subtracting 15 from both sides. And that gives us 35 equals five π‘š. We then divide both sides of this equation by five. And we get seven equals π‘š.

And so if 𝐀 is equal to three 𝐒 minus five 𝐣, 𝐁 is equal to π‘šπ’ plus five 𝐣, and the cross product of 𝐀 and 𝐁 is 50𝐀, π‘š must be equal to seven.

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