Question:
If two vectors \(\vec{A}\) and \(\vec{B}\) are mutually perpendicular, then the component of \(\vec{A}-\vec{B}\) along the direction of \(\vec{A}+\vec{B}\) is:
If two vectors \(\vec{A}\) and \(\vec{B}\) are mutually perpendicular, then the component of \(\vec{A}-\vec{B}\) along the direction of \(\vec{A}+\vec{B}\) is:
Show Hint
For perpendicular vectors,
\[
\vec{A}\cdot\vec{B}=0
\]
which greatly simplifies dot product and magnitude calculations.
Updated On: Jun 22, 2026
- \(\sqrt{|\vec{A}|^2+|\vec{B}|^2}\)
- \(\sqrt{|\vec{A}|^2-|\vec{B}|^2}\)
- \(\dfrac{|\vec{A}|^2-|\vec{B}|^2}{\sqrt{|\vec{A}|^2+|\vec{B}|^2}}\)
- \(\dfrac{|\vec{A}|^2+|\vec{B}|^2}{\sqrt{|\vec{A}|^2-|\vec{B}|^2}}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Formula for component of a vector along another vector.
The component of vector \[ \vec{P} \] along vector \[ \vec{Q} \] is given by \[ \frac{\vec{P}\cdot \vec{Q}}{|\vec{Q}|} \] Here, \[ \vec{P}=\vec{A}-\vec{B} \] and \[ \vec{Q}=\vec{A}+\vec{B} \] Therefore, the required component is \[ \frac{(\vec{A}-\vec{B})\cdot(\vec{A}+\vec{B})}{|\vec{A}+\vec{B}|} \]
Step 2: Evaluate the numerator.
Using distributive property of dot product, \[ (\vec{A}-\vec{B})\cdot(\vec{A}+\vec{B}) \] \[ =\vec{A}\cdot\vec{A}+\vec{A}\cdot\vec{B}-\vec{B}\cdot\vec{A}-\vec{B}\cdot\vec{B} \] Since vectors \(\vec{A}\) and \(\vec{B}\) are mutually perpendicular, \[ \vec{A}\cdot\vec{B}=0 \] Thus, \[ =|\vec{A}|^2-|\vec{B}|^2 \]
Step 3: Evaluate the denominator.
Now, \[ |\vec{A}+\vec{B}| =\sqrt{(\vec{A}+\vec{B})\cdot(\vec{A}+\vec{B})} \] \[ =\sqrt{|\vec{A}|^2+|\vec{B}|^2+2\vec{A}\cdot\vec{B}} \] Again, since \[ \vec{A}\cdot\vec{B}=0, \] we get \[ |\vec{A}+\vec{B}| =\sqrt{|\vec{A}|^2+|\vec{B}|^2} \]
Step 4: Find the required component.
Therefore, \[ \text{Required component} = \frac{|\vec{A}|^2-|\vec{B}|^2} {\sqrt{|\vec{A}|^2+|\vec{B}|^2}} \]
Step 5: Final conclusion.
Hence, the correct answer is \[ \boxed{ \dfrac{|\vec{A}|^2-|\vec{B}|^2} {\sqrt{|\vec{A}|^2+|\vec{B}|^2}} } \]
The component of vector \[ \vec{P} \] along vector \[ \vec{Q} \] is given by \[ \frac{\vec{P}\cdot \vec{Q}}{|\vec{Q}|} \] Here, \[ \vec{P}=\vec{A}-\vec{B} \] and \[ \vec{Q}=\vec{A}+\vec{B} \] Therefore, the required component is \[ \frac{(\vec{A}-\vec{B})\cdot(\vec{A}+\vec{B})}{|\vec{A}+\vec{B}|} \]
Step 2: Evaluate the numerator.
Using distributive property of dot product, \[ (\vec{A}-\vec{B})\cdot(\vec{A}+\vec{B}) \] \[ =\vec{A}\cdot\vec{A}+\vec{A}\cdot\vec{B}-\vec{B}\cdot\vec{A}-\vec{B}\cdot\vec{B} \] Since vectors \(\vec{A}\) and \(\vec{B}\) are mutually perpendicular, \[ \vec{A}\cdot\vec{B}=0 \] Thus, \[ =|\vec{A}|^2-|\vec{B}|^2 \]
Step 3: Evaluate the denominator.
Now, \[ |\vec{A}+\vec{B}| =\sqrt{(\vec{A}+\vec{B})\cdot(\vec{A}+\vec{B})} \] \[ =\sqrt{|\vec{A}|^2+|\vec{B}|^2+2\vec{A}\cdot\vec{B}} \] Again, since \[ \vec{A}\cdot\vec{B}=0, \] we get \[ |\vec{A}+\vec{B}| =\sqrt{|\vec{A}|^2+|\vec{B}|^2} \]
Step 4: Find the required component.
Therefore, \[ \text{Required component} = \frac{|\vec{A}|^2-|\vec{B}|^2} {\sqrt{|\vec{A}|^2+|\vec{B}|^2}} \]
Step 5: Final conclusion.
Hence, the correct answer is \[ \boxed{ \dfrac{|\vec{A}|^2-|\vec{B}|^2} {\sqrt{|\vec{A}|^2+|\vec{B}|^2}} } \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Vectors Questions
- If \[ | \vec{P} + \vec{Q} | = | \vec{P} | = | \vec{Q} |, \] then the angle between \( \vec{P} \) and \( \vec{Q} \) is:
- The angle between two vectors $\vec{i} + \vec{j}$ and $\vec{j} + \vec{k}$ is
- For two vectors $ \mathbf{A} $ and $ \mathbf{B} $, if $ \mathbf{A} + \mathbf{B} = |\mathbf{A} - \mathbf{B}| $, then
- If the angle between two unit vectors $ \vec{A} $ and $ \vec{B} $ is $ \theta $, then \[ |\vec{A} + \vec{B}| = \]
- The sum of the magnitudes of two vectors acting at a point is 18 and the magnitude of their resultant is 12. If the resultant is at \( 90^\circ \) with the vector of smaller magnitude, then the magnitudes of the vectors are
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions