Let vec{a}, vec{b}, vec{c} be three vectors s | vedclass

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

Question

(D) Given $\vec{a} 	imes \vec{b} = 2(\vec{a} 	imes \vec{c})$.This implies $\vec{a} 	imes (\vec{b} - 2\vec{c}) = 0$.Therefore, $\vec{b} - 2\vec{c} =

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given $\vec{a} 	imes \vec{b} = 2(\vec{a} 	imes \vec{c})$.This implies $\vec{a} 	imes (\vec{b} - 2\vec{c}) = 0$.Therefore, $\vec{b} - 2\vec{c} = \lambda \vec{a}$ for some scalar $\lambda$.Taking the magnitude squared on both sides: $|\vec{b} - 2\vec{c}|^2 = \lambda^2 |\vec{a}|^2$.$|\vec{b}|^2 + 4|\vec{c}|^2 - 4(\vec{b} \cdot \vec{c}) = \lambda^2 (1)^2$.Given $|\vec{b}| = 4, |\vec{c}| = 2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos(60^{\circ}) = 4 	imes 2 	imes \frac{1}{2} = 4$.So, $16 + 4(4) - 4(4) = \lambda^2 \Rightarrow \lambda^2 = 16 \Rightarrow \lambda = \pm 4$.Now, $\vec{b} - 2\vec{c} = \pm 4\vec{a}$.Taking the dot product with $\vec{c}$ on both sides:$(\vec{b} - 2\vec{c}) \cdot \vec{c} = \pm 4(\vec{a} \cdot \vec{c})$.$\vec{b} \cdot \vec{c} - 2|\vec{c}|^2 = \pm 4(\vec{a} \cdot \vec{c})$.$4 - 2(4) = \pm 4(\vec{a} \cdot \vec{c}) \Rightarrow -4 = \pm 4(\vec{a} \cdot \vec{c})$.$|\vec{a} \cdot \vec{c}| = |\frac{-4}{\pm 4}| = 1$.

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$. If $|\vec{a}| = 1, |\vec{b}| = 4, |\vec{c}| = 2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|\vec{a} \cdot \vec{c}|$ is:

Similar Questions

If the equation of a straight line passing through a point with position vector $a$ and parallel to the vector $b$ is $r = a + t\,b$,where $t$ is a parameter,then what is the perpendicular distance from this line to a point with position vector $c$?

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane defined by $\hat{i}$ and $\hat{i} + \hat{j}$,and the plane defined by $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$. Find the angle between $\vec{a}$ and $\hat{i} - 2\hat{j} + 2\hat{k}$.

Find a vector that is perpendicular to both vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + \hat{j}$.

If $\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=-3 \hat{i}+5 \hat{j}-4 \hat{k}$ and $\vec{c}=6 \hat{i}-4 \hat{j}+5 \hat{k}$,then $(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=$

The unit vectors orthogonal to $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and coplanar with $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module