Self Studies
Self Studies
Self Studies
Self Studies

Vector Algebra Test - 55

Result Self Studies

Vector Algebra Test - 55
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    If $$\overline {a}. \overline {b}, \overline {c}$$ are unit vectors, then $$|\overline {a} - \overline {b}|^{2} + |\overline {b} - \overline {c}|^{2} + |\overline {c} - \overline {a}|^{2}$$ does not exceed.
    Solution
    $$\textbf{Step1-Simplify the given of expression}$$
                  $$\begin{array}{l} \left| { \vec { a }  } \right| =\left| { \vec { b }  } \right| =\left| { \vec { c }  } \right| =1 \\ { \left| { \vec { a } -\vec { b }  } \right| ^{ 2 } }+{ \left| { \vec { b } -\vec { c }  } \right| ^{ 2 } }+{ \left| { \vec { c } -\vec { a }  } \right| ^{ 2 } }=2\left( { { { \left| a \right|  }^{ 2 } }+{ { \left| b \right|  }^{ 2 } }+{ { \left| c \right|  }^{ 2 } } } \right) -2\left( { \vec { a } \cdot \vec { b } +\vec { b } \cdot \vec { c } +\vec { c } \cdot \vec { a }  } \right)  \\ \left| { \vec { a } +\vec { b } +\vec { c }  } \right| \ge 0 \\ { \left| { \vec { a } +\vec { b } +\vec { c }  } \right| ^{ 2 } }\ge 0 \\ { \left| a \right| ^{ 2 } }+{ \left| b \right| ^{ 2 } }+{ \left| c \right| ^{ 2 } }+2\left( { \vec { a } \cdot \vec { b } +\vec { b } \cdot \vec { c } +\vec { c } \cdot \vec { a }  } \right) \ge 0 \\ \vec { a } \cdot \vec { b } +\vec { b } \cdot \vec { c } +\vec { c } \cdot \vec { a } \ge -\frac { 3 }{ 2 }  \\ 2\left( { \vec { a } \cdot \vec { b } +\vec { b } \cdot \vec { c } +\vec { c } \cdot \vec { a }  } \right) \ge -3 \\ { \left| { \vec { a } -\vec { b }  } \right| ^{ 2 } }+{ \left| { \vec { b } -\vec { c }  } \right| ^{ 2 } }+{ \left| { \vec { c } -\vec { a }  } \right| ^{ 2 } }\ge 2\times 3+3 \\ \ge 9  \end{array}$$
    $$\textbf{Hence , option B is correct}$$
  • Question 2
    1 / -0
    If $$\overrightarrow {a}$$, $$\overrightarrow {b}$$ and $$\overrightarrow {c}$$ be three non-zero vectors, non-coplanar and if $$\overrightarrow {d}$$ is such that $$\bar { a } =\dfrac { 1 }{ y } \left( \overrightarrow { b } +\overrightarrow { c } +\overrightarrow { d}  \right) $$ and  where $$x$$ and $$y$$ are non-zero real numbers, then $$\dfrac { 1 }{ xy } \left( \vec { a } +\vec { b } +\vec { c } +\vec { d }  \right) =$$
    Solution

  • Question 3
    1 / -0
    If the vector $$6\hat { i } -3\hat { j } -6\hat { k } $$ is decomposed into vectors parallel and perpendicular to the vector $$\hat { i } +\hat { j } +\hat { k }$$ then the vectors are :
    Solution
    Given vector : $$6\widehat i - 3\widehat j - 6\widehat k$$
    Let $$6\widehat i - 3\widehat j - 6\widehat k$$ =$$ c+d$$, where c is parallel to $$\widehat i + \widehat j + \widehat k$$ and d is perpendicular to $$\widehat i + \widehat j + \widehat k$$
    hence $$c = m\left( {\widehat i + \widehat j + \widehat k} \right)$$ where m is a scalar
    $$d = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$$ where 
    $$\begin{array} { *{ 20 }{ l } }{ d\cdot \left( { \hat { i } +\hat { j } +\hat { k }  } \right) =0 } \\so\, { { a_{ 1 } }+{ a_{ 2 } }+{ a_{ 3 } }=0 } \\Now\,\, { 6\hat { i } -3\hat { j } -6\hat { k } =m\left( { \hat { i } +\hat { j } +\hat { k }  } \right) +\left( { { a_{ 1 } }\hat { i } +{ a_{ 2 } }\hat { j } +{ a_{ 3 } }\hat { k }  } \right)  } \\so\, equation \,coefficients,\,\, { 6=m+{ a_{ 1 } } } \\ { 3=m+{ a_{ 2 } } } \\ { 6=m+{ a_{ 3 } } } \end{array}$$
    But  $${{a_1} + {a_2} + {a_3} = 0}$$ so $$m =1$$
    then $$\begin{array}{l} { a_{ 1 } }=7;{ a_{ 2 } }=-2;{ a_{ 3 } }=-5 \\ 6\widehat { i } -3\widehat { j } -6\widehat { k } =c+d=m\left( { \widehat { i } +\widehat { j } +\widehat { k }  } \right) +\left( { { a_{ 1 } }\widehat { i } +{ a_{ 2 } }\widehat { j } +{ a_{ 3 } }\widehat { k }  } \right)  \\ =-1\left( { \widehat { i } +\widehat { j } +\widehat { k }  } \right) +\left( { 7\widehat { i } -2\widehat { j } -5\widehat { k }  } \right)  \end{array}$$

  • Question 4
    1 / -0
    If $$|\overline {a}| = 3, |\overline {b}| = 4$$ and $$|\overline {a} - \overline {b}| = 5$$ then $$|a + b| =$$
    Solution
    $$\textbf{Step1-Squaring both side of given expression}$$
                   $$|\vec a-\vec b|^2=(\vec a-\vec b).(\vec a-\vec b)$$
                   $$25=|\vec a|^2+|\vec b|^2-2(\vec a.\vec b)$$
                  $$25=(3)^2+(4)^2-2(\vec a.\vec b)$$
                  $$25=9+16-2(\vec a.\vec b)$$
                  $$\vec a.\vec b=0$$
    $$\textbf{Step2-Find value of}$$ $$|\vec a+\vec b| $$ 
                  $$|\vec a+\vec b|^2=(\vec a+\vec b).(\vec a+\vec b)$$
                  $$=|\vec a|^2+|\vec b|^2+2\vec a.\vec b$$
                  $$=3^2+4^2+2\vec a.\vec b$$
                  $$=9+16+0$$
                  $$=25$$
                  $$|\vec a+\vec b|=5$$
    $$\textbf{Hence , option $$B$$ is the correct answer.}$$
  • Question 5
    1 / -0
    $$\bar { a } ,\bar { c } $$ are unit parallel vectors, $$\left| \bar { b }  \right| =6$$, then $$\bar { b } -3\bar { c } =\lambda \bar { a } $$ if $$\lambda=$$
    Solution

  • Question 6
    1 / -0
    The value of $$\left(a.i\right)i+\left(a.j\right)j+\left(a.k\right)k$$ in terms of vector $$a$$
    Solution
    Let $$\hat{a}={a}_{1}\times\hat{i}+{a}_{2}\times\hat{j}+{a}_{3}\times\hat{k}$$
    $$\therefore \overrightarrow{a}.\hat{i}=\left({a}_{1}\times\hat{i}+{a}_{2}\times\hat{j}+{a}_{3}\times\hat{k}\right).\hat{i}={a}_{1}$$,
    $$ \overrightarrow{a}.\hat{j}={a}_{2},\overrightarrow{a}.\hat{k}={a}_{3}$$
    $$\therefore \left(a.i\right)i+\left(a.j\right)j+\left(a.k\right)k$$
    $$={a}_{1}\times\hat{i}+{a}_{2}\times\hat{j}+{a}_{3}\times\hat{k}=\overrightarrow{a}$$
  • Question 7
    1 / -0
    If the position vectors of A, B, C, D are $$\vec{a}, \vec{b}. 2\vec{a}+3\vec{b}, \vec{a}-2\vec{b}$$ respectively, then $$\vec{AC}, \vec{DB}, \vec{BA}, \vec{DA}$$ are?
    Solution
    Given, 

    $$\vec {OA}=\vec a$$
    $$\vec {OB}=\vec b$$
    $$\vec {OC}=2\vec a+3\vec b$$
    $$\vec {OD}=\vec a-2\vec b$$
    Then, 
    $$\vec {AC}=\vec {OC}-\vec {OA}=\vec a+3\vec b$$
    $$\vec {DB}=\vec {OB}-\vec {OD}=3\vec b-\vec a$$
    $$\vec {BA}=\vec {OA}-\vec {OB}=\vec a-\vec b$$
    $$\vec {DA}=\vec {OA}-\vec {OD}=2\vec b$$

    Therefore, 
    $$\vec {AC},\vec {DB},\vec {BA}$$ and $$\vec {DA}$$ are $$\vec a+3\vec b,3\vec b-\vec a,\vec a-\vec b,2\vec b$$  respectively.

  • Question 8
    1 / -0
    If $$\bar{a}, \bar{b}, \bar{c}$$ are position vectors of the non-collinear points A, B, C respectively, the shortest distance of A and BC is?
    Solution
    vector equation $$\vec{r}=\vec{b}+\lambda(\vec{c}-\vec{b})$$
    Let $$P$$ be the foot of perpendicular of the point $$A$$ on $$BC$$
    Thus, the required distance is $$AP$$.
    $$AP=\sqrt{AB^2-BP^2}$$
    $$AP=|\vec{b}-\vec{a}|^2$$
    $$BP=\cfrac{(\vec{a}-\vec{b}\cdot (\vec{b}-\vec{c})}{|\vec{b}-\vec{c}|}$$
    $$\therefore$$The requiried distance $$AP$$
    $$AP=\sqrt{|\vec{b}-\vec{a}|^2-[\cfrac{(\vec{a}-\vec{b}\cdot (\vec{b}-\vec{c})}{|\vec{b}-\vec{c}|}}]^2$$

  • Question 9
    1 / -0
    What vector must be added to the two vectors $$\hat{i}+2\hat{j}+2\hat{k}$$ and $$2\hat{i}-\hat{j}-\hat{k}$$, so that the resultant may be a unit vector along x-axis
    Solution

  • Question 10
    1 / -0
    If $$A,B,C,D$$ be any four points and $$E$$ and $$F$$ be the mid-points of $$AC$$ and $$BD$$, respectively, then $$\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD}$$ is equal to
    Solution
    Solution:- (C) $$4 \vec{EF}$$
    $$ABCD$$ is a quadrilateral.
    Given that $$F$$ is the mid-point of $$BD$$, we get
    $$\vec{AB} + \vec{AD} = 2 \vec{AF}..... \left( 1 \right)$$
    Similarly $$\vec{CB} + \vec{CD} = 2 \vec{CF} ..... \left( 2 \right) \; \left[ \text{E is the mid point of AC} \right]$$
    Adding $${eq}^{n} \left( 1 \right) \& \left( 2 \right)$$, we have
    $$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = 2 \left( \vec{AF} + \vec{CF} \right)$$
    $$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = 2 \left( \left( -\vec{FA} \right) + \left( - \vec{FC} \right) \right)$$
    $$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = -2 \left( \vec{FA} + \vec{FC} \right)$$
    $$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = -2 \left( 2 \vec{FE} \right)$$
    $$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = -2 \left( -2 \vec{EF} \right)$$
    $$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = 4 \vec{EF}$$

Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now