If b and c are any two non-collinear unit vec | vedclass

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

Question

(A) Let $b$ and $c$ be two non-collinear unit vectors. Since they are non-collinear,$b, c,$ and $k = \frac{b 	imes c}{|b 	imes c|}$ form an orthonor

Vedclass · Accepted Answer

$a$

Vedclass · Answer

(A) Let $b$ and $c$ be two non-collinear unit vectors. Since they are non-collinear,$b, c,$ and $k = \frac{b 	imes c}{|b 	imes c|}$ form an orthonormal basis for the space $\mathbb{R}^3$.Any vector $a$ can be expressed as a linear combination of these basis vectors:$a = (a \cdot b)b + (a \cdot c)c + (a \cdot k)k$Substituting $k = \frac{b 	imes c}{|b 	imes c|}$ into the expression,we get:$a = (a \cdot b)b + (a \cdot c)c + \left(a \cdot \frac{b 	imes c}{|b 	imes c|}ight) \frac{b 	imes c}{|b 	imes c|}$However,the term in the question is $\frac{a \cdot (b 	imes c)}{|b 	imes c|} (b 	imes c)$. Note that $\frac{a \cdot (b 	imes c)}{|b 	imes c|} (b 	imes c) = (a \cdot k) (|b 	imes c| k) = (a \cdot k) (\sin \alpha) k$,where $\alpha$ is the angle between $b$ and $c$. Since $|b 	imes c| = \sin \alpha$,the expression simplifies to:$(a \cdot b)b + (a \cdot c)c + (a \cdot k)k = a$.

If $b$ and $c$ are any two non-collinear unit vectors and $a$ is any vector,then $(a \cdot b)b + (a \cdot c)c + \frac{a \cdot (b \times c)}{|b \times c|} (b \times c) = $

Similar Questions

Let $a$ and $b$ be two unit vectors and $\theta$ is the angle between them. Then,$a+b$ is a unit vector,if

Let $\vec{a}=\hat{i}+2 \hat{j}-2 \hat{k}$ and $\vec{b}=2 \hat{i}-\hat{j}-2 \hat{k}$ be two vectors. If the orthogonal projection vector of $\vec{a}$ on $\vec{b}$ is $\vec{x}$ and the orthogonal projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{y}$, then find $|\vec{x}-\vec{y}|$.

$ABCD$ is a quadrilateral with $\overline{AB}=\bar{a}$,$\overline{AD}=\bar{b}$ and $\overline{AC}=2\bar{a}+3\bar{b}$. If its area is $\alpha$ times the area of the parallelogram with $AB$ and $AD$ as adjacent sides,then the value of $\alpha$ is

If magnitudes of vectors $\vec{a}, \vec{b}, \vec{c}$ are $3, 4,$ and $5$ respectively,and $\vec{a}$ is perpendicular to $\vec{b} + \vec{c}$,$\vec{b}$ is perpendicular to $\vec{c} + \vec{a}$,and $\vec{c}$ is perpendicular to $\vec{a} + \vec{b}$,then find the value of $|\vec{a} + \vec{b} + \vec{c}|$.

If $(a \times b)^{2} + (a \cdot b)^{2} = 144$ and $|a| = 4$,then $|b|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module