Question Video: Finding a Vector given the Results of Its Product by Two Given Vectors Mathematics

If 𝐀 = βˆ’π’ βˆ’ 2𝐣, 𝐁 = βˆ’4𝐒 βˆ’ 4𝐣, 𝐀 Γ— 𝐂 = βˆ’3𝐀, and 𝐂 Γ— 𝐁 = 4𝐀, find 𝐂.

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Video Transcript

If vector 𝐀 is equal to negative 𝐒 minus two 𝐣, vector 𝐁 is equal to negative four 𝐒 minus four 𝐣, the cross product of vector 𝐀 and 𝐂 is negative three 𝐀, and the cross product of vector 𝐂 and vector 𝐁 is four 𝐀, find vector 𝐂.

In this question, we’re given vectors 𝐀 and 𝐁 which lie in the coordinate plane with 𝐒 and 𝐣 as unit vectors. We are also given the cross product of vector 𝐀 and 𝐂 together with the cross product of vector 𝐂 and 𝐁. And we are asked to find vector 𝐂.

We will let this vector be π‘₯𝐒 plus 𝑦𝐣, where π‘₯ and 𝑦 are constants. We recall that if we have two vectors 𝐦 and 𝐧 such that 𝐦 is equal to 𝑒𝐒 plus 𝑓𝐣 and 𝐧 is equal to 𝑔𝐒 plus β„Žπ£, then the cross product of vectors 𝐦 and 𝐧 is equal to the determinant of the two-by-two matrix 𝑒, 𝑓, 𝑔, β„Ž multiplied by the unit vector 𝐀, where the unit vector 𝐀 is perpendicular to the π‘₯𝑦-plane.

Let’s begin by considering the cross product of vectors 𝐀 and 𝐂. Using our general rule, this must be equal to the determinant of the two-by-two matrix negative one, negative two, π‘₯, 𝑦 multiplied by the unit vector 𝐀. The determinant of a matrix is equal to negative one multiplied by 𝑦 minus π‘₯ multiplied by negative two. This is equal to negative 𝑦 plus two π‘₯. So the cross product of vector 𝐀 and vector 𝐂 is negative 𝑦 plus two π‘₯ multiplied by the unit vector 𝐀. We already know that this is equal to negative three 𝐀. We can therefore conclude that negative three is equal to negative 𝑦 plus two π‘₯.

Adding 𝑦 and three to both sides of this equation, we have 𝑦 is equal to two π‘₯ plus three. We will call this equation one and now repeat the process for the cross product of vector 𝐂 and 𝐁. This is equal to the determinant of the two-by-two matrix π‘₯, 𝑦, negative four, negative four multiplied by 𝐀. The determinant of the matrix is π‘₯ multiplied by negative four minus negative four multiplied by 𝑦. This simplifies to negative four π‘₯ plus four 𝑦 such that the cross product of 𝐂 and 𝐁 is negative four π‘₯ plus four 𝑦 multiplied by the unit vector 𝐀. We are told that this is equal to four 𝐀. And negative four π‘₯ plus four 𝑦 is therefore equal to four.

We can divide both sides of this equation by four such that one is equal to negative π‘₯ plus 𝑦. And adding π‘₯ to both sides, we have 𝑦 is equal to π‘₯ plus one. We now have a pair of simultaneous equations we can solve to calculate the values of π‘₯ and 𝑦. Since 𝑦 is equal to two π‘₯ plus three and π‘₯ plus one, then these two expressions must be equal to one another. We can then subtract π‘₯ and three from both sides of the equation. Two π‘₯ minus π‘₯ is π‘₯, and one minus three is negative two, giving us π‘₯ is equal to negative two. We can then substitute this value of π‘₯ into equation one or equation two.

In Equation two, we have 𝑦 is equal to negative two plus one, which gives us 𝑦 is equal to negative one. Vector 𝐂 is therefore equal to negative two 𝐒 minus 𝐣. If 𝐀 is equal to negative 𝐒 minus two 𝐣, 𝐁 is equal to negative four 𝐣 minus four 𝐣, the cross product of 𝐀 and 𝐂 is negative three 𝐀, and the cross product of 𝐂 and 𝐁 is four 𝐀, then vector 𝐂 is equal to negative two 𝐒 minus 𝐣.

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