The dot product of unit vectors hat{n}_1 and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the given vectors be $\vec{A} = 5 \hat{i} + 12 \hat{j}$ and $\vec{B} = 3 \hat{i} + 4 \hat{j}$.First,we find the magnitudes of these vectors:$|

Vedclass · Accepted Answer

$\frac{63}{65}$

Vedclass · Answer

(A) Let the given vectors be $\vec{A} = 5 \hat{i} + 12 \hat{j}$ and $\vec{B} = 3 \hat{i} + 4 \hat{j}$.First,we find the magnitudes of these vectors:$|\vec{A}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.$|\vec{B}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.Now,the unit vectors are $\hat{n}_1 = \frac{\vec{A}}{|\vec{A}|} = \frac{1}{13}(5 \hat{i} + 12 \hat{j})$ and $\hat{n}_2 = \frac{\vec{B}}{|\vec{B}|} = \frac{1}{5}(3 \hat{i} + 4 \hat{j})$.The dot product is $\hat{n}_1 \cdot \hat{n}_2 = \left[ \frac{1}{13}(5 \hat{i} + 12 \hat{j}) ight] \cdot \left[ \frac{1}{5}(3 \hat{i} + 4 \hat{j}) ight]$.$= \frac{1}{65} (5 	imes 3 + 12 	imes 4) = \frac{1}{65} (15 + 48) = \frac{63}{65}$.

The dot product of unit vectors $\hat{n}_1$ and $\hat{n}_2$ that are parallel to $5 \hat{i}+12 \hat{j}$ and $3 \hat{i}+4 \hat{j}$ respectively is

Similar Questions

The area of the parallelogram determined by vectors $A = 2\hat{i} + \hat{j} - 3\hat{k}$ and $B = 12\hat{j} - 2\hat{k}$ is approximately:

The unit vector perpendicular to both vectors $A = -3 \hat{i} - 2 \hat{j} - 3 \hat{k}$ and $B = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}$ is:

Three vectors $\vec{A}, \vec{B}$ and $\vec{C}$ are such that $\vec{A} \cdot \vec{B} = 0$ and $\vec{A} \cdot \vec{C} = 0$. Then $\vec{A}$ is parallel to:

Obtain the scalar product of two mutually perpendicular vectors.

The diagonals of a parallelogram are represented by vectors $\vec{A} = 5\hat{i} - 4\hat{j} + 3\hat{k}$ and $\vec{B} = 3\hat{i} - 2\hat{j} - \hat{k}$. What is the area of the parallelogram (in $\sqrt{3}$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module