Question Video: Finding the Direction Cosines of a Vector | Nagwa Question Video: Finding the Direction Cosines of a Vector | Nagwa

# Question Video: Finding the Direction Cosines of a Vector Mathematics • Third Year of Secondary School

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Find the direction cosines of the vector 𝐀 = 〈5, 2, 8〉.

01:55

### Video Transcript

Find the direction cosines of the vector 𝐀 with components five, two, and eight.

We recall that if vector 𝐯 has components 𝐯 sub 𝑥, 𝐯 sub 𝑦, and 𝐯 sub 𝑧 and direction angles 𝛼, 𝛽, and 𝛾, then the direction cosines cos of 𝛼, cos of 𝛽, and cos of 𝛾 are equal to 𝐯 sub 𝑥 over the magnitude of 𝐯, 𝐯 sub 𝑦 over the magnitude of 𝐯, and 𝐯 sub 𝑧 over the magnitude of 𝐯, respectively, where the magnitude of vector 𝐯 is equal to the square root of 𝐯 sub 𝑥 squared plus 𝐯 sub 𝑦 squared plus 𝐯 sub 𝑧 squared.

In this question, we are told that vector 𝐀 has components five, two, and eight. The magnitude of vector 𝐀 is therefore equal to the square roots of five squared plus two squared plus eight squared. Five squared is equal to 25, two squared is equal to four, and eight squared is equal to 64. Summing these three values gives us an answer of 93. Therefore, the magnitude of vector 𝐀 is the square root of 93. This means that the cos of angle 𝛼 is equal to five over the square root of 93, the cos of angle 𝛽 is equal to two over the square root of 93, and, finally, the cos of angle 𝛾 is equal to eight over the square root of 93. The direction cosines of vector 𝐀 are five over the square root of 93, two over the square root of 93, and eight over the square root of 93.

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