Tag: vectors from a geometric viewpoint

Questions Related to vectors from a geometric viewpoint

If three vectors $\overline{a},\overline{b},\ \overline{c}$ are such that $\overline{a}\neq 0$, $\overline{a}\times\overline{b}=2\overline{a}\times\overline{c},\ |\overline{a}|=|\overline{c}|=1,\ |\overline{b}|=4$ and the angle between $|\overline{b}|$ and $|\overline{c}|$ is $\displaystyle \cos^{-1}\frac{1}{4}$, then $\overline{b}-2\overline{c}=\lambda\overline{a}$ where $\lambda$ is equal to

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\pm 2$

  2. $\pm 4$

  3. $\displaystyle \dfrac{1}{2}$

  4. $\displaystyle \dfrac{1}{4}$

Show answer
Correct Option: B

If $\vec{a}\times\vec{b}=\vec{c}\times\vec{d}$ and $\vec{a}\times\vec{c}=\vec{b}\times\vec{d}$, then

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\vec{a}+\vec{b}=\vec{c}+\vec{d}$

  2. $\vec{a}-\vec{d}$ is parallel to $\vec{b}-\vec{c}$

  3. $\vec{a}-\vec{d}$ is perpendicular to $\vec{b}-\vec{c}$

  4. $\vec{a}-\vec{b}$ is perpendicular to $\vec{a}-\vec{b}$

Show answer
Correct Option: B
Explanation:

$\vec{a}\times \vec{b}= \vec{c}\times \vec{d}$
$\vec{a}\times \vec{c}= \vec{b}\times \vec{d}$
$\vec{a}\times (\vec{b}-\vec{c})= (\vec{c}-\vec{b})\times \vec{d}$
$(\vec{a}-\vec{d})\times (\vec{b}-\vec{c})= 0$
$(\vec{a}-\vec{d})\ \parallel \ (\vec{b}-\vec{c})$

If $\vec {a},\vec {b},\ \vec {c}$ are non-zero non-collinear vectors such that $\vec {a}\times\vec {b}=\vec {b}\times\vec {c}=\vec {c}\times\vec {a}$ , then $\vec {a}+\vec {b}+\vec {c}=$

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $abc$

  2. $-1$

  3. $\vec {0}$

  4. $2$

Show answer
Correct Option: C
Explanation:

$\vec{a}\times \vec{b}=\vec{b}\times \vec{c}$
$(\vec{a}+\vec{c})\times \vec{b}=o$
$(\vec{a+\vec{c}}+\vec{b})\times \vec{b}=\vec{b\times \vec{b}}$
$(\vec{a}+\vec{c}+\vec{b})\times \vec{b}=0$
$\vec{a}+\vec{c}+\vec{b}=\vec{0}$   hence $\vec{a},\vec{b},\vec{c}$ are non collinear

If $\vec {a}\times \vec {b}=\vec {c}\times \vec {d},\vec {a}\times \vec {c}=\vec {b}\times \vec {d}$, then

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\vec {a}-\vec {d}$ is parallel to $\vec {b}-\vec {c}$

  2. $\vec {a}-\vec {b}$ is parallel to $\vec {c}-\vec {d}$

  3. $\vec {a}-\vec {c}$ is parallel to $\vec {b}-\vec {d}$

  4. $\vec {a}+\vec {b}$ is parallel to $\vec {c}+\vec {d}$

Show answer
Correct Option: A
Explanation:

$\vec{a}\times \vec{b}=\vec{c}\times \vec{d}  -(1)$
$\vec{a}\times \vec{c}=\vec{b}\times \vec{d}  -(2)$
$(1) - (2)$
$\vec{a}\times \vec{b}+\vec{c}\times \vec{a}=\vec{c}\times \vec{d}+\vec{d}\times \vec{b}$
$(\vec{a}-\vec{d})\times \vec{b}=c\times (\vec{d}-\vec{a})$
$(\vec{a}-\vec{d})\times \vec{b}+(\vec{d}-\vec{a})\times \vec{c}=0$
$(\vec{a}-\vec{d})\times (\vec{b}-\vec{c})=0$
which mean $(\vec{a}-\vec{d})||(\vec{b}-\vec{c})$

If $\vec {a}$ and $\vec {b}$ are not perpendicular to each other and $\vec {r}\times\vec {a}=\vec {b}\times\vec {a},\ \vec {r}.\vec {c}=0$, then $\vec {r}$ is equal to

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\vec {a}-\vec {c}$

  2. $\vec {b}+\lambda\vec {a}$, for all scalars $\lambda$

  3. $\displaystyle \vec {b}-\dfrac{(\vec {b}.\vec {c})}{(\vec {a}.\vec {c})}\vec {a}$

  4. $\vec {a}+\vec {c}$

Show answer
Correct Option: C
Explanation:

Let $r,a,b$ and $c$ be vectors.
It is given that
$r\times a=b\times a$
$r\times a-(b\times a)=0$
$(r-b)\times a=0$
Hence $r-b$ is a vector parallel to vector $a$.
$r=b+\mu a$ ...(i)
It is given that $r.c=0$.
Hence $r$ vector is perpendicular to $c$ vector.
$(b+\mu a).c=0$ ...(from i)
$b.c+\mu(a.c)=0$
$\mu(a.c)=-b.c$
$\mu=-\dfrac{b.c}{a.c}$
Hence $r=b-\dfrac{b.c}{a.c}a$

If $a$ and $b$ are two unit vectors inclined at an angle $\dfrac { \pi  }{ 3 }$, then $\left{ a\times \left( b+a\times b \right)  \right} \cdot b$ is equal to

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\dfrac { 1 }{ 4 } $

  2. $\dfrac { -3 }{ 4 } $

  3. $\dfrac { 3 }{ 4 } $

  4. $\dfrac { 1 }{ 2 } $

Show answer
Correct Option: B
Explanation:

Given, $\left| a \right| =\left| b \right| =1$ and $a\cdot b=\cos { \dfrac { \pi  }{ 3 }  } $
Consider,
$\left{ a\times \left( b+a\times b \right)  \right} \cdot b=\left{ a\times b+a\times \left( a\times b \right)  \right} \cdot b$
            $=\left( a\times b \right) \cdot b+\left{ \left( a\cdot b \right) \cdot a-\left( a\cdot a \right) \cdot b \right} \cdot b$
            $=\left[ \begin{matrix} a & b & b \end{matrix} \right] +{ \left( a\cdot b \right)  }^{ 2 }-{ \left| a \right|  }^{ 2 }{ \left| b \right|  }^{ 2 }$
            $=0+\cos ^{ 2 }{ \dfrac { \pi  }{ 3 }  } -1=\dfrac { 1 }{ 4 } -1=\dfrac { -3 }{ 4 } $

Let $\vec{\lambda }=\vec{a}\times \left ( \vec{b}+\vec{c} \right )$, $\vec{\mu }=\vec{b}\times \left ( \vec{c}+\vec{a} \right )$ and $\vec{\nu }=\vec{c}\times \left ( \vec{a}+\vec{b} \right )$, then

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\vec{\lambda }+\vec{\mu }=\vec{\nu }$

  2. $\vec{\lambda }, \vec{\mu }, \vec{\nu }$ are coplanar

  3. $\vec{\lambda }+\vec{\nu }=2\vec{\mu }$

  4. None of these

Show answer
Correct Option: B
Explanation:

$\vec { \lambda  } +\vec { \mu  } =\vec { a } \times \left( \vec { b } +\vec { c }  \right) +\vec { b } \times \left( \vec { c } +\vec { a }  \right) $

$= \vec { a } \times \vec { b } +\vec { a } \times \vec { c } +\vec { b } \times \vec { c } +\vec { b } \times \vec { a }$
$ = \left( \vec { a } +\vec { b }  \right) \times \vec { c }$
$= -\vec { \nu  } \ \Rightarrow \quad \vec { \nu  }$
$ =-\left( \vec { \lambda  } +\vec { \mu  }  \right) $

one vector is expressed as linear combination of other two vectors
Hence,
$\vec { \lambda  } ,\vec { \mu  } ,\vec { \nu  } $ are coplanar vectors.

Let $\vec{r}\times \vec{a}=\vec{b}\times \vec{a}$ and $\vec{r}.\vec{c}=0$, where $\vec{a}\vec{b}\neq 0$, then $\vec{r}$ is equal to

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\vec{b}+t\vec{a}$ where $t$ is a scalar

  2. $\displaystyle \vec{b}-\dfrac{\vec{b}.\vec{c}}{\vec{a}.\vec{c}}\vec{a}$

  3. $\vec{a}-\vec{c}$

  4. $None\ of\ these$

Show answer
Correct Option: B
Explanation:

$\vec{r}\times \vec{a}=\vec{b}\times \vec{a}$

$\Rightarrow \vec{r}\times \vec{a}-\vec{b}\times \vec{a}=0$
$\Rightarrow (\vec{r}-\vec{b}\times \vec{a} = 0\Rightarrow \vec{r} = \vec{b}+t\vec{a}$

Now taking dot product with $\vec{c}$ both side,
$\Rightarrow \vec{r}\cdot\vec{c} = \vec{b}\cdot\vec{c}+t \vec{a}\cdot\vec{c}=0$
$\Rightarrow t = -\dfrac{\vec{b}\cdot\vec{c}}{\vec{a}\cdot\vec{c}}$

Hence $\vec{r} = \vec{b}-\dfrac{\vec{b}\cdot\vec{c}}{\vec{a}\cdot\vec{c}}\vec{a}$

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are any three vectors in space then $\left ( \overrightarrow{c}+\overrightarrow{b} \right )\times \left ( \overrightarrow{c}+\overrightarrow{a} \right ).\left ( \overrightarrow{c}+\overrightarrow{b}+\overrightarrow{a} \right )$ is equal to

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $3\begin{bmatrix}

    \overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c}

    \end{bmatrix}$

  2. $0$

  3. $\begin{bmatrix}

    \overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c}

    \end{bmatrix}$

  4. None of these

Show answer
Correct Option: C
Explanation:

$\left( \overrightarrow{c}+\overrightarrow{b} \right )\times \left ( \overrightarrow{c}+\overrightarrow{a} \right ).\left ( \overrightarrow{c}+\overrightarrow{b}+\overrightarrow{a} \right)$

$= \left(\overrightarrow{c} \times \overrightarrow{a}+\overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{b}\times \overrightarrow{a}\right)\cdot \left ( \overrightarrow{c}+\overrightarrow{b}+\overrightarrow{a} \right)$
$= \overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{b} \times \overrightarrow{c}\cdot \overrightarrow{c} + \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{c}$
$+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b} \times \overrightarrow{c}\cdot \overrightarrow{b} + \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{b}$
$+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{a}+\overrightarrow{b} \times \overrightarrow{c}\cdot \overrightarrow{a} + \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{a}$
$=0+0+ \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{b}+0 + 0+0+\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a} + 0$
$= \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}$