If $\vec {a} = \hat {i} + \hat{j} + \hat{k}$, $\ve
Question:

If a=i^+j^+k^\vec {a} = \hat {i} + \hat{j} + \hat{k}, b=i+3j+5k^\vec {b} = \vec {i} + 3\vec {j} + 5\hat{k} and c=7i^+9j^+11k^\vec {c} = 7\hat{i} + 9\hat {j} + 11 \hat{k}, then the area of Parallelogram having diagonals a+b\vec{a} +\vec{b} and b+c\vec{b} +\vec{c} is

Updated On: Apr 15, 2024
  • 464 \sqrt{6} s units
  • 1221\frac{1}{2} \sqrt{21} s units
  • 4848
  • 6 \sqrt{6} s units
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

a=i+j+k,b=i+3j+5ka = i + j + k, b = i + 3j + 5k
and c=7i+9j+11kc = 7i + 9j + 11k
Let A=a+bA = a + b
=(i+j+k)+(i+3j+5k)= (i + j + k) + (i + 3j + 5k)
=2i+4j+6k= 2i + 4j + 6k
and B=b+cB = b + c
=(i+3j+5k)+(7i+9j+11k)= (i + 3j +5k) + (7i + 9j + 11k)
=8i+12j+16k= 8i + 12j + 16k
\therefore Area of parallelogram
=12A×B= \frac{1}{2}\left|A \times B\right|
(  A\because \,\,A and BB are diagonals)
ijk24681216 \begin{vmatrix}i&j&k\\ 2&4&6\\ 8&12&16\end{vmatrix}
=12i(6472)j(3248)+k(2432)= \frac{1}{2} \left|i\left(64-72\right) - j \left(32 - 48\right) + k \left(24 - 32\right)\right|
=128i+6j8k= \frac{1}{2} \left| - 8i + 6j - 8 k\right|
=(4)2+(8)2+(4)2= \sqrt{\left(-4\right)^{2} + \left(8\right)^{2}+\left(-4\right)^{2}}
=96= \sqrt{96}
=46= 4\sqrt{6} sq units
Was this answer helpful?
0
0

Top Questions on Vector Algebra

View More Questions

Concepts Used:

Vector Algebra

A vector is an object which has both magnitudes and direction. It is usually represented by an arrow which shows the direction(→) and its length shows the magnitude. The arrow which indicates the vector has an arrowhead and its opposite end is the tail. It is denoted as

The magnitude of the vector is represented as |V|. Two vectors are said to be equal if they have equal magnitudes and equal direction.

Vector Algebra Operations:

Arithmetic operations such as addition, subtraction, multiplication on vectors. However, in the case of multiplication, vectors have two terminologies, such as dot product and cross product.