$\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{B} = 5\hat{i} - 4\hat{k}$
The magnitude of area of the parallelogram is ________ (Rounded off to two decimal places)
The area of a parallelogram formed by two adjacent vectors $\vec{A}$ and $\vec{B}$ is equal to the magnitude of their cross product, denoted as $|\vec{A} \times \vec{B}|$.
Given vectors are:
The cross product $\vec{A} \times \vec{B}$ is calculated using a determinant:
$ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 5 & 0 & -4 \end{vmatrix} $Expanding the determinant:
$ = \hat{i}((3)(-4) - (-1)(0)) - \hat{j}((2)(-4) - (-1)(5)) + \hat{k}((2)(0) - (3)(5)) $ $ = \hat{i}(-12 - 0) - \hat{j}(-8 - (-5)) + \hat{k}(0 - 15) $ $ = -12\hat{i} - \hat{j}(-3) - 15\hat{k} $ $ = -12\hat{i} + 3\hat{j} - 15\hat{k} $The resulting vector is $-12\hat{i} + 3\hat{j} - 15\hat{k}$. The magnitude of this vector is calculated as:
$ |\vec{A} \times \vec{B}| = \sqrt{(-12)^2 + (3)^2 + (-15)^2} $ $ = \sqrt{144 + 9 + 225} $ $ = \sqrt{378} $Calculating the square root:
$ \sqrt{378} \approx 19.44155 $Rounding off to two decimal places gives $19.44$.
The magnitude of the area of the parallelogram is approximately 19.44.
What is the length of projection of the vector \(\rm \hat{i}+2 \hat{j}+3 \hat{k}\) on the vector \(\rm2 \hat{i}+3 \hat{j}-2 \hat{k}\) ?
Consider the following in respect of the vectors \(\rm \vec{a}=(0,1,1)\) and \(\rm \vec{b}=(1,0,1) \) :
1. The number of unit vectors perpendicular to both \(\rm \vec{a}\) and \(\rm \vec{b}\) is only one.
2. The angle between the vectors is \(\frac{\pi}{3}\).
Which of the statements given above is/are correct?
Consider the following points :
1. (-1, -3, 1)
2. (-1, 3, 2)
3. (-2, 5, 3)
Which of the above points lie on the line joining A and B ?
What is the magnitude of \(\overrightarrow{A B}\) ?
If \({\rm{\vec d}} = {\rm{x\hat i}} + {\rm{y\hat j}} + {\rm{z\hat k}}\) , then which of the following equations is/are correct?
1. y – x = 4
2. 2z – 3 = 0
Select the correct answer using the code given below: