Vectors problem|How to solve vector in physics| how To prove; (I) A. [ B × C ] = B. [ C × A ] = C. [ A × B ], B. [ A × C ] = -A. [ B × C ] (III) and A. [ B × A ] = 0 is correct

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

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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