If the sum of two unit vectors is a unit vect | vedclass

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

Question

(C) Let $\vec{a}$ and $\vec{b}$ be two unit vectors, so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Given that their sum is a unit vector, $|\vec{a} + \vec{b

Vedclass · Accepted Answer

$\sqrt{3}$ units

Vedclass · Answer

(C) Let $\vec{a}$ and $\vec{b}$ be two unit vectors, so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Given that their sum is a unit vector, $|\vec{a} + \vec{b}| = 1$. Squaring both sides, we get $|\vec{a} + \vec{b}|^2 = 1^2$. Using the identity $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$, we have $1 + 1 + 2(\vec{a} \cdot \vec{b}) = 1$. This simplifies to $2 + 2(\vec{a} \cdot \vec{b}) = 1$, which gives $2(\vec{a} \cdot \vec{b}) = -1$. Now, we need to find the magnitude of their difference, $|\vec{a} - \vec{b}|$. We know that $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$. Substituting the known values, $|\vec{a} - \vec{b}|^2 = 1 + 1 - (-1) = 1 + 1 + 1 = 3$. Therefore, $|\vec{a} - \vec{b}| = \sqrt{3}$ units.

If the sum of two unit vectors is a unit vector, then the magnitude of their difference is

Similar Questions

Find the position vector of the midpoint of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.

If $a = \alpha \hat{i} + 3 \hat{j} - 6 \hat{k}$ and $b = 2 \hat{i} - \hat{j} + \beta \hat{k}$,then the values of $\alpha, \beta$ so that $a$ and $b$ may be collinear are

Consider two points $P$ and $Q$ with position vectors $\overrightarrow{OP} = 3\vec{a} - 2\vec{b}$ and $\overrightarrow{OQ} = \vec{a} + \vec{b}$. Find the position vector of a point $R$ which divides the line segment joining $P$ and $Q$ in the ratio $2:1$ internally.

If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ are parallel vectors,then which of the following is true?

The direction cosines of the vector $\hat{i}-2\hat{j}+3\hat{k}$ are . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module