The projection of the vector vec{a} = hat{i} | vedclass

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

Question

(B) The projection of a vector $\vec{a}$ on a vector $\vec{b}$ is given by the formula: $	ext{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.

Vedclass · Accepted Answer

$\frac{19}{9}$

Vedclass · Answer

(B) The projection of a vector $\vec{a}$ on a vector $\vec{b}$ is given by the formula: $	ext{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.Given $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = 4\hat{i} - 4\hat{j} + 7\hat{k}$.First,calculate the dot product $\vec{a} \cdot \vec{b} = (1)(4) + (-2)(-4) + (1)(7) = 4 + 8 + 7 = 19$.Next,calculate the magnitude of vector $\vec{b}$:$|\vec{b}| = \sqrt{4^2 + (-4)^2 + 7^2} = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$.Therefore,the projection is $\frac{19}{9}$.

The projection of the vector $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ on the vector $\vec{b} = 4\hat{i} - 4\hat{j} + 7\hat{k}$ is:

Similar Questions

For a triangle $ABC$ with vertices $A(1, 0, 0)$,$B(0, 1, 0)$,and $C(0, 0, 1)$,the angle $A = \dots$

Let $a$ and $b$ be two unit vectors and $\theta$ is the angle between them. Then,$a+b$ is a unit vector,if

Let $a, b, c$ be the position vectors of the vertices of a triangle $ABC$. The vector area of triangle $ABC$ is

Show that the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$ form the vertices of a right-angled triangle.

If the position vectors of the vertices of a triangle are $2 \hat{i}-\hat{j}+\hat{k}$,$\hat{i}-3 \hat{j}-5 \hat{k}$,and $3 \hat{i}-4 \hat{j}-4 \hat{k}$,then the triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module