AP EAPCET 04-Jul-2022 Shift 2 Paper

(c) In option (a)
(A. A) (B. C)
(A2)(BC‌cos‌θ)
=A2BC‌cos‌θ Scalar value
In option (b)
(A×B)(B×C)‌=(ABsin‌θ‌

∧

n

)⋅(BCsin‌φ‌k)
‌=(ABsin‌(θ)(BCsin‌φ)(‌

∧

n

⋅

∧

k

)
where, ‌

∧

n

is unit vector perpendicular to both vector A and B and hat‌k is unit vector perpendicular to both vector B and C.
∴‌

∧

n

⋅k= scalar value
∴(A×C)⋅(B×C) is a scalar value
In option (c)
‌(A×C)×(B×C)=(ACsin‌θ1)‌

∧

r

×(BCsin‌θ2)‌

∧

r

2..
‌=(ACsin‌θ1)(BCsin‌θ2)(‌

∧

r

×‌

∧

r

2)
‌

∧

r

1×‌

∧

r

2 is a vector product, hence it gives a vector value.
Therefore, (A×C)×(B×C) is a vector value.
In option (d)
A×(B×C) is vector triple product which gives a vector value.