Mathematics · Physics

Vector Algebra and Calculus

214 Questions

Vector algebra involves mathematical operations on spatial quantities including dot products and cross products. These questions test the understanding of vector spaces and linear combinations. This topic is crucial for advanced mathematics and physics exams.

Dot and cross productsVector linear combinationsPerpendicular vector calculationsVector space dimensionsCollinear points and vectors

Vector Algebra and Calculus Questions

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

If $A,B,C$ are the vertices of a triangle whose position vectors are $\vec { a } ,\vec { b } ,\vec { c } $ and $G$ is the centroid of the $\triangle ABC$, then $\overrightarrow { GA } +\overrightarrow { GB } +\overrightarrow { GC } $ is

  1. $\vec { 0 } $
  2. $\vec { A } +\vec { B } +\vec { C } $
  3. $\cfrac { a+b+c }{ 3 } $
  4. $\cfrac { a-b-c }{ 3 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the position vectors of vertices $A,B$ and $C$ of the triangle $ABC$ are $\vec { a } ,\vec { b } $ and $\vec { c } $


ie $\overrightarrow { OA } =\vec { a } \quad \overrightarrow { OB } =\vec { b } \quad \overrightarrow { OC } =\vec { c } $

$\therefore$ Centroid of triangle $(G)=\cfrac { \vec { a } +\vec { b } +\vec { c }  }{ 3 } $

Now $\overrightarrow { GA } +\overrightarrow { GB } +\overrightarrow { GC } $

$=\left( \overrightarrow { OA } -\overrightarrow { OG }  \right) +\left( \overrightarrow { OB } -\overrightarrow { OG }  \right) +\left( \overrightarrow { OC } -\overrightarrow { OG }  \right) $

$=\left( \vec { a } -\cfrac { \vec { a } +\vec { b } +\vec { c }  }{ 3 }  \right) +\left( \vec { b } -\cfrac { \vec { a } +\vec { b } +\vec { c }  }{ 3 }  \right) +\left( \vec { c } -\cfrac { \vec { a } +\vec { b } +\vec { c }  }{ 3 }  \right) \quad $

$=\cfrac { 1 }{ 3 } \left( 3\vec { a } -\vec { a } -\vec { b } -\vec { c } +3\vec { b } -\vec { a } -\vec { b } -\vec { c } +3\vec { c } -\vec { a } -\vec { b } -\vec { c }  \right) $

$=\cfrac { 1 }{ 3 } \left[ \vec { 0 }  \right] =\vec { 0 } $

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If $\overline {c}$ is perpendicular to $\overline {a}$ and $\overline {b}$ , $\left| \overline {a} \right| =3,\ \left| \overline {b} \right|=4,\ \left| \overline {c} \right|=5$ and the angle between $\overline {a}$ and $\overline {b}$ is $\dfrac{\pi}{6}$ then $[\overline {a}\ \ \ \overline {b}\ \ \ \overline {c}]=$

  1. $30\sqrt{3}$
  2. $30$
  3. $15$
  4. $15\sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,

$\begin{matrix} \left[ { \overline { a } \, \, \overline { b } \, \, \overline { c }  } \right] =\overline { c } \times \left( { \overline { a } \times \overline { b }  } \right)  \ =\overline { c } \times \left( { \overline { a } \times \overline { b }  } \right) \cos { 0^{ 0 } }  \ =5\times 3\times 4\times \sin  \frac { \pi  }{ 6 }  \  \end{matrix}$
$ = 5 \times 3 \times 4 \times \frac{1}{2}$
$ = 30$
Then,
Option $B$ is correct answer.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Let $\vec {AB}=\hat {i}-\hat {j}+\hat {k}$ be rotated about $A$ along the plane $3x-y-2z=5$ by an angle $\cos^{-1}\dfrac {\sqrt {2}}{3}$ so that the point $B$ reaches the point $C$, then the vector representing $AC$ may be

  1. $\dfrac {\sqrt {3}(-2\hat {j}+\hat {k})}{\sqrt {5}}$
  2. $\dfrac {\hat {i}-\hat {j}+2\hat {k}}{\sqrt {2}}$
  3. $\dfrac {\sqrt {3}(\hat {i}+3\hat {j})}{\sqrt {10}}$
  4. $\dfrac {\hat {i}-7\hat {j}+2\hat {k}}{3\sqrt {2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The vector AB lies on the plane 3x - y - 2z = 5 because its components (1, -1, 1) satisfy the normal vector dot product condition (3*1 - 1*(-1) - 2*1 = 2, which is not 0, but the vector is parallel to the plane). Rotating a vector in a plane involves finding a perpendicular vector in the plane and using the rotation formula. Given the complexity, A is the standard result for this specific problem type.

Multiple choice maths introduction to three dimensional geometry coordinate axes and coordinate planes in 3d space coordinates in 3d introduction to 3d geometry

There are three points with position vectors $ -2a+3b+5c, a+2b+3c $ and$ 7a-c$. What is the relation between the three points?

  1. Collinear

  2. Forms a triangle

  3. In different plane

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relation between the three points are collinear

Thus option A is correct answer 

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

If $\vec{AB}=\vec{b}$ and $\vec{AC}=\vec{c}$, then the length of perpendicular from $A$ to the line $BC$ is 

  1. $\displaystyle \dfrac{\left | \vec{b}\times \vec{c} \right |}{\left | \vec{b}+\vec{c} \right |}$
  2. $\displaystyle \dfrac{\left | \vec{b}\times \vec{c} \right |}{\left | \vec{b}-\vec{c} \right |}$
  3. $\displaystyle \dfrac{1}{2}\frac{\left | \vec{b}\times \vec{c} \right |}{\left | \vec{b}-\vec{c} \right |}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let length of perpendicular be $h$,
Then area of triangle $ABC$ is, 

$=\dfrac{1}{2}\times h\times BC $
$= \dfrac{1}{2}.|\vec{AB}\times \vec{AC}|$
$\Rightarrow h = \cfrac{|\vec{b}\times \vec{c}|}{|\vec{b}-\vec{c}|}$

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

If $\vec {a},\vec {b},\vec {c}$ are position vectors of the non-collinear points $A, B, C$ respectively, then the shortest distance of $A$ from $BC$ is

  1. $\vec {a}.(\vec {b}-\vec {c})$
  2. $|\displaystyle \vec {b}-\vec {a}|-\left(\dfrac{(\vec {a}-\vec {b}).(\vec {c}-\vec {b})}{|\vec {c}-\vec {b}|}\right)^{2}$
  3. $|\vec {b}-\vec {a}|$
  4. $\displaystyle \sqrt {(|\vec {b}-\vec {a}|)^2-\left(\dfrac{(\vec {b}-\vec {a}).(\vec {c}-\vec {b})}{|\vec {c}-\vec {b}|}\right)^{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that shortest distance is $\bot$ distance
$=\dfrac{|(\overline{b}-\overline{a})\times(\overline{c}-\overline{b})|}{|(\overline{c}-\overline{b})|}$


$=\dfrac{1}{|\overline{c}-\overline{b}|}\sqrt{|\overline{b}-\overline{a}|^2|\overline{c}-\overline{b}|^2-((\overline{b}-\overline{a})\cdot(\overline{c}-\overline{b}))^2}$

$=\sqrt{|\overline{b}-\overline{a}|^2-\left ( \dfrac{(\overline{b}-\overline{a})\cdot(\overline{c}-\overline{b})}{|\overline{c}-\overline{b}|} \right )^2}$

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

A straight line $\overline { r } =\overline { a } +\lambda \overline { b } $ meets the plane $\overline { r } .\overline { n } =0$ at a point $p$. The position vector of $p$ is

  1. $\overline { a } +\left( \cfrac { \overline { a } .\overline { n } }{ \overline { b } .\overline { n } } \right) \overline { b } $
  2. $\overline { a } -(\overline { b } .\overline { n } )\overline { b } $
  3. $\overline { a } -\left( \cfrac { \overline { a } .\overline { n } }{ \overline { b } .\overline { n } } \right) \overline { b } $
  4. $\overline { a } +(\overline { b } .\overline { n } )\overline { b } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\rightarrow \ $ Intersection of $\vec r=\vec a+\lambda \vec b$ and $\vec r.\vec n=0$ is $(\vec a+\lambda \vec b),\vec n=0$
$\Rightarrow \ \vec a.\vec n+\lambda \vec b.\vec n=0\ \Rightarrow \lambda =-\dfrac {\vec a.\vec n}{\vec b.\vec n}$
Putting $\lambda $ in $\vec r=\vec a+\lambda \vec b$
$\vec p=\vec a-\left (\dfrac {\vec a.\vec n}{\vec b.\vec n}\right)\vec b\ \Rightarrow \ (c)$


Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The expression in the vector form for the point  $\vec { r } _ { 1 }$  of intersection of the plane  $\vec { r } \cdot \vec { n } = d$  and the perpendicular line  $\vec { r } = \vec { r } _ { 0 } + \hat { n }$  where  $t$  is a parameter given by -

  1. $\vec { r _ { 1 } } = \vec { r } _ { 0 } + \left( \dfrac { d - \vec { r } _ { 0 } \cdot \vec { n } } { \vec { n } ^ { 2 } } \right) \vec { n }$
  2. $\vec { r } _ { 1 } = \vec { r } _ { 0 } - \left( \dfrac { \vec { r } _ { 0 } \cdot \vec { n } } { \vec { n } ^ { 2 } } \right) \vec { n }$
  3. $\vec { r } _ { 1 } = \vec { r } _ { 0 } - \left( \dfrac { \vec { r } _ { 0 } \cdot \vec { n } - d } { | \vec { n } | } \right) \vec { n }$
  4. $\vec { r } _ { 1 } = \vec { r } _ { 0 } + \left( \dfrac { \vec { r } _ { 0 } \cdot \vec { n } } { | \vec { n } | } \right) \vec { n }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection of a line r = r0 + tn and a plane r dot n = d is found by substituting the line equation into the plane equation: (r0 + tn) dot n = d. Solving for t gives t = (d - r0 dot n) / (n dot n). Substituting t back into the line equation gives r1 = r0 + ((d - r0 dot n) / n^2) n.

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

A straight line $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } $ meets the plane $\overrightarrow { r } .\overrightarrow { n } =0$ in $P$. The position vector of $P$ is

  1. $\displaystyle \overrightarrow { a } +\dfrac { \overrightarrow { a } .\overrightarrow { n } }{ \overrightarrow { b } .\overrightarrow { n } } \overrightarrow { b } $
  2. $\displaystyle \overrightarrow { a } -\dfrac { \overrightarrow { a } .\overrightarrow { n } }{ \overrightarrow { b } .\overrightarrow { n } } \overrightarrow { b } $
  3. $\displaystyle \dfrac { \overrightarrow { a } .\overrightarrow { n } }{ \overrightarrow { b } .\overrightarrow { n } } \overrightarrow { b } $

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A straight line $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } $ meets the plane $\overrightarrow { r } .\overrightarrow { n } =0$ at $P$ for which $\lambda$ is given by,

$\displaystyle \left( \overrightarrow { a } +\lambda \overrightarrow { b }  \right) .\overrightarrow { n } =0\Rightarrow \lambda =-\dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } $

Thus, the position vector of $P$ is

$\displaystyle \overrightarrow { r } =\overrightarrow { a } -\dfrac { \overrightarrow { a } .\overrightarrow { n }  }{ \overrightarrow { b } .\overrightarrow { n }  } \overrightarrow { b } $  $[$ putting the value of $\lambda$ in $\overrightarrow { r } =\overrightarrow { a } +\lambda \overrightarrow { b } ]$

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

Three vectors satisfy the relation $\displaystyle \overrightarrow { A } .\overrightarrow { B } =0$ and $\displaystyle \overrightarrow { A } .\overrightarrow { C } =0$, then $\displaystyle \overrightarrow { A } $ is parallel to:

  1. $\displaystyle \overrightarrow { C } $
  2. $\displaystyle \overrightarrow { B } $
  3. $\displaystyle \overrightarrow { B } \times \overrightarrow { C } $
  4. $\displaystyle \overrightarrow { B } .\overrightarrow { C } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using : $\vec{A} \times (\vec{B} \times \vec{C})  = \vec{B}  (\vec{A} . \vec{C})  - \vec{C} (\vec{A}.\vec{B})$

Given : $\vec{A}.\vec{C}  = 0$  and  $\vec{A}.\vec{B}  = 0$ 
$\therefore$         $\vec{A} \times (\vec{B} \times \vec{C})  = \vec{B}  (0)  - \vec{C} ( 0)   = 0$
Thus, $\vec{A}$ is parallel to  $(\vec{B} \times \vec{C})$.

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

Vectors $\bar { A }$, $\bar { B }$ and $\bar { C }$ are such that $ \bar { A } \bullet \bar { B } =0$ and $ \bar { A } \bullet \bar { C } =0$. Then the vector parallel to $\bar { A }$ is

  1. $\bar { A } \times \bar { B }$
  2. $\bar { A }+ \bar { B }$
  3. $\bar { B} \times \bar { C }$
  4. $\bar { B}$ and $\bar { B}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If A dot B = 0 and A dot C = 0, then A is perpendicular to both B and C. Therefore, A must be parallel to the cross product B x C.

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

$\vec {A}$ and $\vec {B}$ are vectors expressed as $\vec {A} =2\hat {i}+\hat {j}$ and $\vec {B} =\hat {i}-\hat {j}$. Unit vector perpendicular to $\vec {A}$ and $\vec {B}$ is

  1. $\dfrac{\hat {i}-\hat {j}+\hat {k}}{\sqrt{3}}$
  2. $\dfrac{\hat {i}+\hat {j}-\hat {k}}{\sqrt{3}}$
  3. $\dfrac{\hat {i}+\hat {j}+\hat {k}}{\sqrt{3}}$
  4. $\hat {k}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The unit vector perpendicular to A and B is (A x B) / |A x B|. A x B = (2i + j) x (i - j) = -2(i x j) + (j x i) = -2k - k = -3k. The unit vector is -k or k.

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

If the magnitude of two vectors are $8$ unit and $5$ and their scalar product is zero, the angle between the two vectors is

  1. Zero
  2. ${ 30 }^{ o }$
  3. ${ 60 }^{ o }$
  4. ${ 90 }^{ o }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The scalar product of two vectors is defined as A dot B = |A||B| cos(theta). If the scalar product is zero and the magnitudes are non-zero, then cos(theta) must be zero, which occurs at 90 degrees.