AP EAMCET · Maths · Vector Algebra

If $A B C D E F$ is a regular hexagon with $\mathbf{A B}=\mathbf{a}$ and $\mathbf{B C}=\mathbf{b}$, then $\mathbf{C E}$ equals to

A $\mathbf{b}-\mathbf{a}$
B $-\mathbf{b}$
C $\mathbf{b}-2 \mathbf{a}$
D $\mathbf{a}-2 \mathbf{b}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\mathbf{b}-2 \mathbf{a}$

Step-by-step Solution

Detailed explanation

Given, $A B C D E F$ is a regular hexagon $\mathbf{A B}=\mathbf{a}, \mathbf{B C}=\mathbf{b}$…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From AP EAMCET

Match the items of List-

I

with the items of List-

II

and choose the correct option

List- $I$		List- $II$
$(A)$	If $A$ is a non-singular matrix of order $3$ and $\| A \| = a,$ then $\|{(adj A^{- 1})}^{- 1}\| =$	$(I)$	null matrix
$(B)$	A is a non-singular matrix of order $3$ and $B$ is any matrix of order $3$ such that $A B = O$ , then $B$ is	$(II)$	$a^{2}$
$(C)$	$\|\begin{matrix} 1 & x & x^{2} \\ \cos (a - b) y & \cos α y & \cos (a + b) y \\ \sin (a - b) y & \sin α y & \sin (a + b) y \end{matrix}\|$ does not depend on	$(III)$	$b$
$(D)$	$A$ is a square matrix of order $3$ and $B = A - A^{T}$ , then $\|B\|$ is	$(IV)$	$a$
		$(V)$	$0$

The correct answer is

AP EAMCET 2019 Medium

Which of the following will form an ideal solution?

AP EAMCET 2021 Easy

Explore more questions on app

If \(A B C D E F\) is a regular hexagon with \(\mathbf{A B}=\mathbf{a}\) and \(\mathbf{B C}=\mathbf{b}\), then \(\mathbf{C E}\) equals to

Answer & Solution

Step-by-step Solution

See the Complete Solution