If overrightarrow{u}=overrightarrow{a}-overri | vedclass

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

Question

(A) Given $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$.We calculate the c

Vedclass · Accepted Answer

$2 \sqrt{16-(\overrightarrow{a} \cdot \overrightarrow{b})^2}$

Vedclass · Answer

(A) Given $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$.We calculate the cross product: $\overrightarrow{u} 	imes \overrightarrow{v} = (\overrightarrow{a}-\overrightarrow{b}) 	imes (\overrightarrow{a}+\overrightarrow{b})$.Using the distributive property of the cross product: $\overrightarrow{u} 	imes \overrightarrow{v} = \overrightarrow{a} 	imes \overrightarrow{a} + \overrightarrow{a} 	imes \overrightarrow{b} - \overrightarrow{b} 	imes \overrightarrow{a} - \overrightarrow{b} 	imes \overrightarrow{b}$.Since $\overrightarrow{a} 	imes \overrightarrow{a} = 0$ and $\overrightarrow{b} 	imes \overrightarrow{b} = 0$, and $\overrightarrow{b} 	imes \overrightarrow{a} = -(\overrightarrow{a} 	imes \overrightarrow{b})$, we get:$\overrightarrow{u} 	imes \overrightarrow{v} = 0 + \overrightarrow{a} 	imes \overrightarrow{b} + \overrightarrow{a} 	imes \overrightarrow{b} - 0 = 2(\overrightarrow{a} 	imes \overrightarrow{b})$.Taking the magnitude: $|\overrightarrow{u} 	imes \overrightarrow{v}| = 2|\overrightarrow{a} 	imes \overrightarrow{b}|$.Using the identity $|\overrightarrow{a} 	imes \overrightarrow{b}|^2 = |\overrightarrow{a}|^2 |\overrightarrow{b}|^2 - (\overrightarrow{a} \cdot \overrightarrow{b})^2$:$|\overrightarrow{a} 	imes \overrightarrow{b}| = \sqrt{|\overrightarrow{a}|^2 |\overrightarrow{b}|^2 - (\overrightarrow{a} \cdot \overrightarrow{b})^2}$.Given $|\overrightarrow{a}| = 2$ and $|\overrightarrow{b}| = 2$, then $|\overrightarrow{a}|^2 = 4$ and $|\overrightarrow{b}|^2 = 4$.Thus, $|\overrightarrow{a} 	imes \overrightarrow{b}| = \sqrt{4 	imes 4 - (\overrightarrow{a} \cdot \overrightarrow{b})^2} = \sqrt{16 - (\overrightarrow{a} \cdot \overrightarrow{b})^2}$.Therefore, $|\overrightarrow{u} 	imes \overrightarrow{v}| = 2 \sqrt{16 - (\overrightarrow{a} \cdot \overrightarrow{b})^2}$.

If $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$, $\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$, $|\overrightarrow{a}|=|\overrightarrow{b}|=2$, then $|\overrightarrow{u} \times \overrightarrow{v}|$ is equal to

Similar Questions

If $\vec{a}, \vec{b}, \text{ and } \vec{c}$ are unit vectors,then $|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2$ does not exceed:

The major product of the following reaction is:

The de-Broglie wavelength of an electron having $80 eV$ energy is nearly ($1 eV = 1.6 \times 10^{-19} J$,Mass of the electron $= 9 \times 10^{-31} kg$,Planck's constant $= 6.6 \times 10^{-34} J-s$) (in $\text{ Å}$)

Heat energy is incident on the surface at the rate of $1000 ~J/min$. If the coefficient of absorption is $0.8$ and the coefficient of reflection is $0.1$,then the heat energy transmitted by the surface in $5 ~min$ is: (in $~J$)

If,in a $\triangle ABC$,$\tan \frac{A}{2} = \frac{5}{6}$ and $\tan \frac{C}{2} = \frac{2}{5}$,then $a, b, c$ are such that :

Vedclass Test Series

Exam Paper Generator

Online Exam Module