If two vectors A and B are mutually perpendic | vedclass

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

Question

(C) Given that vectors $A$ and $B$ are mutually perpendicular,so $A \cdot B = 0$.We need to find the component of vector $(A-B)$ along the direction o

Vedclass · Accepted Answer

$\frac{|A|^2-|B|^2}{\sqrt{|A|^2+|B|^2}}$

Vedclass · Answer

(C) Given that vectors $A$ and $B$ are mutually perpendicular,so $A \cdot B = 0$.We need to find the component of vector $(A-B)$ along the direction of vector $(A+B)$.The component of a vector $P$ along the direction of vector $Q$ is given by the formula: $	ext{Component} = \frac{P \cdot Q}{|Q|}$.Here,$P = A-B$ and $Q = A+B$.First,calculate the magnitude of $(A+B)$:$|A+B| = \sqrt{(A+B) \cdot (A+B)} = \sqrt{A \cdot A + B \cdot B + 2(A \cdot B)} = \sqrt{|A|^2 + |B|^2 + 0} = \sqrt{|A|^2 + |B|^2}$.Now,calculate the dot product $(A-B) \cdot (A+B)$:$(A-B) \cdot (A+B) = A \cdot A + A \cdot B - B \cdot A - B \cdot B = |A|^2 + 0 - 0 - |B|^2 = |A|^2 - |B|^2$.Finally,the component is:$	ext{Component} = \frac{(A-B) \cdot (A+B)}{|A+B|} = \frac{|A|^2 - |B|^2}{\sqrt{|A|^2 + |B|^2}}$.

Column $I$	Column $II$
$(A) (A+B)$	$(p)$ North-east
$(B) (A-B)$	$(q)$ Vertically upwards
$(C) (A \times B)$	$(r)$ Vertically downwards
$(D) (A \times B) \times (A \times B)$	$(s)$ None

If two vectors $A$ and $B$ are mutually perpendicular,then the component of $A-B$ along the direction of $A+B$ is

Similar Questions

The area of the parallelogram represented by the vectors $\overrightarrow A = 2\hat i + 3\hat j$ and $\overrightarrow B = \hat i + 4\hat j$ is ....... $units^2$.

Consider two vectors $\vec{F}_1 = 2\hat{i} + 5\hat{k}$ and $\vec{F}_2 = 3\hat{j} + 4\hat{k}$. The magnitude of the scalar product of these vectors is

Vector $A$ is pointing eastwards and vector $B$ is pointing northwards. Match the following two columns:

Column $I$ Column $II$

$(A) (A+B)$ $(p)$ North-east

$(B) (A-B)$ $(q)$ Vertically upwards

$(C) (A \times B)$ $(r)$ Vertically downwards

$(D) (A \times B) \times (A \times B)$ $(s)$ None

Two vectors $a \hat{i} + b \hat{j} + \hat{k}$ and $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ are perpendicular to each other. Given $3a + 2b = 7$,the ratio of $a$ to $b$ is $\frac{x}{2}$. The value of $x$ is:

If $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{B} \times \overrightarrow{A}$,then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module