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Question Video: Using the Norm of Vectors to Find the Value of an Unknown Mathematics

Given that 𝐀 = (6, 4, π‘˜) and |𝐀| = 2√(17) units, determine the possible values of π‘˜.

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Video Transcript

Given that vector 𝐀 is equal to six, four, π‘˜ and the magnitude of vector 𝐀 is equal to two root 17 units, determine the possible values of π‘˜.

Let’s begin by considering the general vector 𝐏 with coordinates π‘₯, 𝑦, and 𝑧. The magnitude of vector 𝐏 is equal to the square root of π‘₯ squared plus 𝑦 squared plus 𝑧 squared. This means that if vector 𝐀 has coordinates six, four, and π‘˜, then the magnitude of vector 𝐀 is equal to the square root of six squared plus four squared plus π‘˜ squared. We know that the magnitude in this question is equal to two root 17. Two root 17 is equal to the square root of six squared plus four squared plus π‘˜ squared.

In order to solve this equation for π‘˜, our first step is to square both sides. The right-hand side becomes six squared plus four squared plus π‘˜ squared. The left-hand side becomes four multiplied by 17, as two squared is equal to four and root 17 squared is equal to 17. Four multiplied by 17 is equal to 68. Six squared equals 36. And four squared equals 16. The equation simplifies to 68 equals 36 plus 16 plus π‘˜ squared. 36 plus 16 is equal to 52. Subtracting 52 from both sides of the equation gives us 68 minus 52 is equal to π‘˜ squared. This means that π‘˜ squared is equal to 16. Our final step is to square root both sides of the equation. The square root of 16 is equal to positive four or negative four. The square root of π‘˜ squared is equal to π‘˜. This means that the two possible values of π‘˜ are four and negative four.

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