Solving vector problem

Let’s solve maths(vectors)

Use A = 2i + 3j – 5k

B = 3i + j +2k

C = I – j + 3k

To prove;

(I) A. [ B × C ] = B. [ C × A ] = C. [ A × B ]

(II) B. [ A × C ] = -A. [ B × C ]

(III) A. [ B × A ] = 0 is correct

SOLUTIONS

Note: The coefficient of I, j and k are the ones multiplying themselves the I, j and k still remain the same. I.e alike factors multiply themselves

(I) A. [ B × C ] = B. [ C × A ] = C. [ A × B ]

first we’ll solve for A. [ B × C ] then relate it to B. [ C × A ] after we’ll check if it’s true for C. [ A × B ]

Let’s get down to business

Solve for A. [ B × C ]

2i+3j-5k[ (3i +j+2k) × (i-j+3k) ]

2i+3j-5k[ 3i×i +j×-j + 2k×3k ]

2i+3j-5k[ 3i -j+6k ]

2i × 3i + 3j×-j – 5k×6k

=6i -3j -30k

For B. (C × A)

3i+j+2k[ (i – j+3k) × (2i+3j-5k) ]

3i+j+2k[ 2i-3j-15k ]

3i×2i + j×-3j + 2k×-15k

=6i -3j -30k

For C. [A × B]

i – j +3k [ (2i +3j -5k) × (3i + j +2k) ]

i – j +3k [ 2i×3i +3j×j – 5k×2k ]

i – j +3k [ 6i +3j – 10k ]

i×6i – j×3j + 3k×-10k

= 6i -3j -30k

So therefore, A. [ B × C ] = B. [ C × A ] = C. [ A × B]

(2) ANSWER FOR QUESTION 2

(II) B. [ A × C ] = -A. [ B × C ]

Solve first for B. [ A × C ]

3i + j + 2k [ (2i + 3j – 5k) × (i – j + 3k) ]

3i + j + 2k [ 2i – 3j – 15k ]

3i×2i + j×-3j + 2k×-15k

= 6i -3j -30k

Then solve for right hand side

-A. (B × C)

-[2i+3j-5k] [ (3i +j+2k) × (i-j+3k) ]

-2i-3j+5k[ 3i×i +j×-j + 2k×3k ]

-2i-3j+5k[ 3i -j+6k ]

-2i × 3i – 3j×-j + 5k×6k

=-6i +3j + 30k

So therefore B. [ A × C ] is not equal to -A. [ B × C ]

(3) ANSWER FOR QUESTION 3

(III) A. [ B × A ] = 0 is correct

As, A×B = ABsin theta

ABsin theta is a vector which is perpendicular to the plane having A vector and V vector which implies that it is also perpendicular to A vector.

And A = AB cos theta

NOTE: An angle is perpendicular only if it is 90°

Therefore,

A. [ B× A ] = AB cos theta. × AB son theta

Theta = 90°

= AB cos 90° × AB son 90°

= AB[0] × AB [1]

Anything multiply by zero is what

Answer: sure it will be zero, so therefore it’s true

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