Let vec{a}, vec{b}, and vec{c} be three unit | vedclass

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

Question

(A) Given that $\vec{a}, \vec{b},$ and $\vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1.$Using the vector triple product formula,$

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) Given that $\vec{a}, \vec{b},$ and $\vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1.$Using the vector triple product formula,$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}.$Given $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{1}{2} \vec{b},$ we have $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{1}{2} \vec{b}.$Since $\vec{b}$ and $\vec{c}$ are non-parallel,they are linearly independent. Comparing the coefficients of $\vec{b}$ and $\vec{c},$ we get $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = 0.$We know $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \alpha = 1 \cdot 1 \cdot \cos \alpha = 0 \implies \alpha = 90^{\circ}.$We know $\vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \beta = 1 \cdot 1 \cdot \cos \beta = \frac{1}{2} \implies \beta = 60^{\circ}.$Thus,$|\alpha - \beta| = |90^{\circ} - 60^{\circ}| = 30^{\circ}.$

Similar Questions

If $\bar{a}=\hat{j}-\hat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\hat{k}$,then the vector $\bar{b}$ satisfying $\bar{a} \times \bar{b}+\bar{c}=\vec{0}$ and $\bar{a} \cdot \bar{b}=3$ is

The vector $\vec{a}=-\hat{i}+2 \hat{j}+\hat{k}$ is rotated through a right angle,passing through the $y$-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3 \vec{a}+\sqrt{2} \vec{b}$ on $\vec{c}=5 \hat{i}+4 \hat{j}+3 \hat{k}$ is

If $\vec{a}$ is a vector such that $\vec{a} \times \hat{i}=\hat{j}+\hat{k}$ and $\vec{a} \cdot \hat{i}=1$, then the equation of the line passing through the point $\hat{i}+\hat{j}+\hat{k}$ and parallel to $\vec{a}$ is

Let $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{v}=-3 \hat{j}+2 \hat{k}$ be vectors in $R^3$ and $\vec{w}$ be a unit vector in the $XY$-plane. Then,the maximum value of $|(\vec{u} \times \vec{v}) \cdot \vec{w}|$ is:

The area of the parallelogram whose diagonals are $a = 3i + j - 2k$ and $b = i - 3j + 4k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module