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Area of Trapezium by Division into Shapes of Known Area

What is the a...

Question

What is the area of the rectangle having vertices A, B, C and D with position vectors $-^i + \frac{1}{2}^j + 4^k,^i + \frac{1}{2}^j + 4^k,^i - \frac{1}{2}^j + 4^k$ and $-^i - \frac{1}{2}^j + 4^k$ ?

1/2 square unit

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1 square unit

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2 square unit

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4 square unit

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Solution

The correct option is C 2 square unit
From figure we can see,
$A B = \sqrt{2^{2} + 0^{2} + 0^{2}} = 2$
$B C = \sqrt{0^{2} + 1^{2} + 0^{2}} = 1$
So, area $= A B \times B C$
$= 2$
Hence, C is correct.

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Similar questions

Q. Area of a rectangle having vertices

A, B, C

and

D

with position vectors

-^i + \frac{1}{2}^j + 4^k

^i + \frac{1}{2}^j + 4^k

^i - \frac{1}{2}^j + 4^k

and

-^i - \frac{1}{2}^j + 4^k

Area of a rectangle having vertices A, B, C, and D with position vectors and respectively is

(A) (B) 1

Area of a rectangle having vertices $A (-^i + \frac{1}{2}^j + 4^k), B (^i + \frac{1}{2}^j + 4^k), C (^i - \frac{1}{2}^j + 4^k)$ and $D (-^i - \frac{1}{2}^j + 4^k)$ is

a) $\frac{1}{2}$
b) 1
c) 2
d) 4

Q. At what value of the parameter n the vectors

\to a =^i + 2^j + 4^k

and

\to a =^i + 2^j + 2 n^k

are equal

Q. Let

\to a =^i + 2^j + 4^k, \to b =^i + λ^j + 4^k

and

\to c = 2^i + 4^j + (λ^{2} - 1)^k

be coplanar vectors. Then the non-zero vector

\to a \times \to c

is :

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What is the area of the rectangle having vertices A, B, C and D with position vectors −^i+12^j+4^k,^i+12^j+4^k,^i−12^j+4^k and −^i−12^j+4^k ?

The correct option is C 2 square unitFrom figure we can see,AB=√22+02+02=2 BC=√02+12+02=1So, area =AB×BC =2Hence, C is correct.

What is the area of the rectangle having vertices A, B, C and D with position vectors $-^i + \frac{1}{2}^j + 4^k,^i + \frac{1}{2}^j + 4^k,^i - \frac{1}{2}^j + 4^k$ and $-^i - \frac{1}{2}^j + 4^k$ ?

The correct option is C 2 square unit
From figure we can see,
$A B = \sqrt{2^{2} + 0^{2} + 0^{2}} = 2$
$B C = \sqrt{0^{2} + 1^{2} + 0^{2}} = 1$
So, area $= A B \times B C$
$= 2$
Hence, C is correct.