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If a= i+j-2k then the direction cosines are

User Heberda
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Explanation:

Direction Cosine (l) = a_x / |a|

Direction Cosine (m) = a_y / |a|

Direction Cosine (n) = a_z / |a|

Where:

(a_x, a_y, a_z) are the components of the vector.

|a| is the magnitude of the vector.

In your case, you have the vector a = i + j - 2k, and its components are:

a_x = 1

a_y = 1

a_z = -2

To find the magnitude of the vector |a|, you can use the formula:

|a| = sqrt(a_x^2 + a_y^2 + a_z^2)

|a| = sqrt(1^2 + 1^2 + (-2)^2)

|a| = sqrt(1 + 1 + 4)

|a| = sqrt(6)

Now, you can find the direction cosines:

Direction Cosine (l) = a_x / |a| = 1 / sqrt(6)

Direction Cosine (m) = a_y / |a| = 1 / sqrt(6)

Direction Cosine (n) = a_z / |a| = -2 / sqrt(6)

So, the direction cosines of the vector a = i + j - 2k are:

l = 1 / sqrt(6)

m = 1 / sqrt(6)

n = -2 / sqrt(6)

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