# If →A,→B,→C Are Mutually Perpendicular Vectors of Equal Magnitudes, Show that the Vector →A+ →B+→C is Equally Inclined to →A,→B and →C. - Mathematics

Sum

If veca, vecb, vecc are mutually perpendicular vectors of equal magnitudes, show that the vector veca +  vecb+ vecc is equally inclined to veca, vecb and vecc.

#### Solution

Since veca , vecb  "and"  vec c are mutually perpendicular vectors, we have

veca . vecb = vecb . vecc = vec c . vec a = 0

It is given that:

|vec a| = |vec b| = |vec c|

Let vector vec a + vec b + vec c be inclined to veca , vecb  "and"  vec c at angles theta_1,theta_2 and theta_3 respectively.

Then, we have:

cos θ1 = ((vec a + vec b + vec c).vec a)/(|vec a + vec b + vec c| |vec a|) = (vec a .vec a + vec b. vec a + vec c.veca)/(|vec a + vec b + vec c||vec a|)

= |vec a|^2/(|vec a + vec b + vec c||vec a|)         .....[vecb. vec a = vec c. vec a = 0]

=|veca|/(|veca + vecb + vecc|)

cosθ2 = ((vec a+vec b +vec c).vec b)/(|vec a + vec b + vec c|.|vec b|) =(vec a .vec b + vec b. vec b + vec c.vec b)/(|vec a + vec b + vec c||vec b|)

= |vec b|^2/(|vec a + vec b + vec c||vec b|)         .....[vec a. vec b = vec c. vec b = 0]

=|vecb|/(|veca + vecb + vecc|)

cosθ3 = ((vec a+vec b +vec c).vec c)/(|vec a + vec b + vec c|.|vec c|) =(vec a .vec c + vec b. vec c + vec c.vec c)/(|vec a + vec b + vec c||vec c|)

= |vec c|^2/(|vec a + vec b + vec c||vec c|)         .....[vec a. vec c = vec b. vec c = 0]

=|vecc|/(|veca + vecb + vecc|)

Now, as |vec a| = |vec b| = |vec c| , cos θ1 = cosθ2 = cosθ3

∴ θ1 = θ2 = θ3

Hence, the vector (veca + vecb + vecc) is equally inclined to veca, vecb " and " vecc.

Concept: Magnitude and Direction of a Vector
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#### APPEARS IN

NCERT Class 12 Maths
Chapter 10 Vector Algebra
Q 14 | Page 458
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