Given that vector 𝐀 is equal to negative six, three, negative two and vector 𝐁 is equal to four, one, six, determine the component of vector 𝐀 along vector 𝐁.
We begin by recalling that we can find the component of some vector 𝐀 in the direction of another vector 𝐁 using scalar projection, where the scalar projection of vector 𝐀 onto vector 𝐁 is equal to the dot product of vector 𝐀 and vector 𝐁 divided by the magnitude of vector 𝐁. We can find the dot product of two vectors by multiplying their corresponding components and then finding the sum of these values. In this question, we have negative six multiplied by four plus three multiplied by one plus negative two multiplied by six. This is equal to negative 24 plus three minus 12, which is equal to negative 33.
The magnitude of vector 𝐁 is equal to the square root of the sum of the squares of its individual components. We have the square root of four squared plus one squared plus six squared. This is equal to root 53. The component of vector 𝐀 along vector 𝐁 is therefore equal to negative 33 over root 53. We can simplify this by rationalizing the denominator. We multiply the numerator and denominator by root 53. This gives us a final answer of negative 33 root 53 over 53. This is the component of vector 𝐀 along vector 𝐁.