Question

Find the vector of magnitude 9 which is equally inclined to the coordinate axes

Answer 1

Answer:

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Explanation:

Let R be the vector. Since it is equally inclined to the co-ordinate axes, all the direction angles would be equal. i.e. A=B=C

We know that the sum of square of direction cosine is equal to 1. So,

(cosA)^2+(cosB)^2+(cosC)^2=1

3*(cosA)^2=1 ( because A=B=C)

cosA=cosB=cosC= 1/sqrt3 & cosA=cosB=cosC= -1/sqrt3

So, the unit vector along the vector R would be:

(1/sqrt3)i+(1/sqrt3)j+(1/sqrt3)k & -(1/sqrt3)i-(1/sqrt3)j-(1/sqrt3)k

1/sqrt3(i+j+k) & -1/sqrt3(i+j+k)

Because the magnitude of R is 9

So,

Vector R=9*1/sqrt3(i+j+k) & R=9*(-1/sqrt3(i+j+k))

R= 3sqrt3(i+j+k) & R= -3sqrt3(i+j+k)

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Answer 2

Let R be the vector. Since it is equally inclined to the co-ordinate axes, all the direction angles would be equal. i.e. A=B=C

We know that the sum of square of direction cosine is equal to 1. So,

(cosA)^2+(cosB)^2+(cosC)^2=1

3*(cosA)^2=1 ( because A=B=C)

cosA=cosB=cosC= 1/sqrt3 & cosA=cosB=cosC= -1/sqrt3

So, the unit vector along the vector R would be:

(1/sqrt3)i+(1/sqrt3)j+(1/sqrt3)k & -(1/sqrt3)i-(1/sqrt3)j-(1/sqrt3)k

1/sqrt3(i+j+k) & -1/sqrt3(i+j+k)

Bcause the magnitude of R is 9

Vector R=9*1/sqrt3(i+j+k) & R=9*(-1/sqrt3(i+j+k))

R= 3sqrt3(i+j+k) & R= -3sqrt3(i+j+k)