If vec{a} and vec{b} are two vectors such tha | vedclass

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

Question

(B) Given that $|\vec{a}| = |\vec{b}| = \sqrt{14}$ and $\vec{a} \cdot \vec{b} = -7$. We know that $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(B) Given that $|\vec{a}| = |\vec{b}| = \sqrt{14}$ and $\vec{a} \cdot \vec{b} = -7$. We know that $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta$. Substituting the values: $-7 = (\sqrt{14})(\sqrt{14}) \cos 	heta = 14 \cos 	heta$. Thus,$\cos 	heta = -\frac{7}{14} = -\frac{1}{2}$. Since $\cos 	heta = -\frac{1}{2}$,we have $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - (-\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$. Now,the expression is $\frac{|\vec{a} 	imes \vec{b}|}{|\vec{a} \cdot \vec{b}|} = \frac{|\vec{a}| |\vec{b}| \sin 	heta}{|\vec{a}| |\vec{b}| |\cos 	heta|} = \frac{\sin 	heta}{|\cos 	heta|}$. Substituting the values: $\frac{\frac{\sqrt{3}}{2}}{|-\frac{1}{2}|} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$.

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{14}$ and $\vec{a} \cdot \vec{b}=-7$,then $\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}=$

Similar Questions

$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = \dots$

If the moduli of the vectors $a, b, c$ are $3, 4, 5$ respectively and $a$ and $b + c$,$b$ and $c + a$,$c$ and $a + b$ are mutually perpendicular,then the modulus of $a + b + c$ is

$a, b$ and $c$ are three vectors with magnitudes $|a| = 4, |b| = 4, |c| = 2$ such that $a$ is perpendicular to $(b + c)$,$b$ is perpendicular to $(c + a)$,and $c$ is perpendicular to $(a + b)$. It follows that $|a + b + c|$ is equal to

In a triangle $ABC$,if $|\overrightarrow{BC}|=3$,$|\overrightarrow{AC}|=5$,and $|\overrightarrow{BA}|=7$,then the projection of the vector $\overrightarrow{BA}$ on $\overrightarrow{BC}$ is equal to:

$A$ unit vector in the $xy$-plane which is perpendicular to $4i - 3j + k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module