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Vector Algebra Test - 30

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Vector Algebra Test - 30
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  • Question 1
    1 / -0
    If $$A, B$$ are two points on the curve $$y=x^{2}$$ in the $$x-y$$ plane satisfying $$\vec{OA}.\hat{i}=1$$ and $$\vec{OB}.\hat{i}=-2$$ then the length of the vectors $$2\vec{OA}-3\vec{OB}$$ is
    Solution
    Let  $$\vec{OA}=x_1 i+x_1^2j$$ and $$\vec{OB}=x_2i+x_2^2j$$
    Now $$\vec{OA}\cdot i = x_1 = 1$$ (given)
    and $$\vec{OB}\cdot \vec{i} =x_2 =-2$$ (given)
    So, $$\vec{OA}= i+j$$ and $$\vec{OB}=-2i+4j$$
    Thus $$2\vec{OA}-3\vec{OB}=2i+2j-(-6i+12j)=8i-10j$$
    $$\therefore |2\vec{OA}-3\vec{OB}|=\sqrt{8^2+(-10)^2}=\sqrt{164}=2\sqrt{41}$$
  • Question 2
    1 / -0
    In a tetrahedron if two pairs of opposite edges are at a right angles then the third pair is inclined at an angle of
    Solution
    Let the position vectors of the vertices be$$ \overline o, \overline a, \overline b\  and\ \overline c. $$
    One opposite pair= $$\overline a.(\overline b- \overline c) = 0$$
    hence,$$ a.b = a.c$$
    Similarly from other pair $$b.a = b.c$$
    Hence , from these 2 equations, $$a.b = b.c = a.c $$
    Hence $$c.(a-b) = a.c - b.c = 0$$
    Hence proved

  • Question 3
    1 / -0
    If $$\overrightarrow { a } .\overrightarrow { b } =0$$ and also $$\overrightarrow { a } \times \overrightarrow { b } =0,$$ then
    Solution
    $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ cannot be perpendicular both at the same time.
    So in order to hold $$\overrightarrow { a } .\overrightarrow { b } =0$$ and $$\overrightarrow { a }\times\overrightarrow { b } =0$$, simultaneously, either $$\overrightarrow { a } $$ or $$\overrightarrow{b}$$ must be zero.
  • Question 4
    1 / -0
    Let $$G$$ be the centroid of $$\triangle ABC$$. If $$\overrightarrow{AB}=\overrightarrow{a}$$ and $$\overrightarrow{AC}=\overrightarrow{b},$$ then $$\overrightarrow{AG}$$, in terms of $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ is
    Solution
    Let $$A$$ be the origin.
    then $$\overrightarrow { AB } =\overrightarrow { a } ,\overrightarrow { AC } =\overrightarrow { b } $$ implies that the position vectors of $$B$$ and $$C$$ are $$\overrightarrow{b}$$ and $$\overrightarrow{c}$$ respectively.
    Let $$AD$$ be the median and $$G$$ be the centroid. Then,
    $$P.V.$$ of $$\displaystyle D=\dfrac { \overrightarrow { a } +\overrightarrow { b }  }{ 2 } $$,
    $$P.V.$$ of $$\displaystyle G=\dfrac { \overrightarrow { a } +\overrightarrow { b }  }{ 3 } $$
    $$\displaystyle \therefore \overrightarrow { AG } =\dfrac { \overrightarrow { a } +\overrightarrow { b }  }{ 3 } $$

  • Question 5
    1 / -0
    If $$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $$ are the position vectors of the vertices of an equilateral triangle whose orthocentre is at the origin, then the centroid of the triangle satisfies which of the following relation?
    Solution
    The position vector of the centroid of the triangle is $$\displaystyle \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c }  }{ 3 } $$
    Since the triangle is equilateral, therefore the orthocenter coincides with the centroid and hence $$\displaystyle \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c }  }{ 3 } =0$$
    $$\therefore \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =0$$
  • Question 6
    1 / -0
    If $$\alpha(\vec a \times \vec b)+\beta(\vec b \times \vec c)+\gamma(\vec c \times \vec {a})=\vec{0}$$ and at least one of the scalars $$\alpha,\ \beta,\gamma$$ is non-zero, then the vectors $$\vec{a},\vec{b},\vec{c}$$ are
    Solution
    As per the question, and the definition of linearly dependent vectors, we get that $$(\overline a \times \overline b), (\overline a \times \overline c)$$ and $$(\overline b \times \overline c)$$ are linearly dependent.
    Now, since they are linearly dependent, any one of them can be written in terms of other two and thus we get that they are coplanar.
  • Question 7
    1 / -0
    If $$I$$ is the center of a circle inscribed in a triangle $$ABC$$, then $$|BC|IA+|CA|IB+|AB|IC$$
    Solution
    We know that the position vaector of the incenter of a $$\triangle ABC$$ is given by

    $$\displaystyle \dfrac { BC.\overrightarrow { a } +CA.\overrightarrow { b } +AB.\overrightarrow { c }  }{ BC+CA+AB } $$

    where $$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $$ are the affixes of the vertices of the triangle

    $$\therefore$$, in the given triangle $$ABC$$, the affix of $$I$$ is given by

    $$\displaystyle \dfrac { \left| \overrightarrow { BC }  \right| \overrightarrow { IA } +\left| \overrightarrow { CA }  \right| \overrightarrow { IB } +\left| \overrightarrow { AB }  \right| \overrightarrow { IC }  }{ \left| \overrightarrow { BC }  \right| +\left| \overrightarrow { CA }  \right| +\left| \overrightarrow { AB }  \right|  } $$

    But is the position vector of $$I$$ with respect to $$I$$.

    $$\displaystyle \therefore \overrightarrow { 0 } =\dfrac { \left| \overrightarrow { BC }  \right| \overrightarrow { IA } +\left| \overrightarrow { CA }  \right| \overrightarrow { IB } +\left| \overrightarrow { AB }  \right| \overrightarrow { IC }  }{ \left| \overrightarrow { BC }  \right| +\left| \overrightarrow { CA }  \right| +\left| \overrightarrow { AB }  \right|  } $$

    $$\displaystyle \Rightarrow \left| \overrightarrow { BC }  \right| \overrightarrow { IA } +\left| \overrightarrow { CA }  \right| \overrightarrow { IB } +\left| \overrightarrow { AB }  \right| \overrightarrow { IC } =\overrightarrow { 0 } $$
  • Question 8
    1 / -0
    lf the four points $$\overline{a},\overline{b},\overline{c},\overline{d}$$ are coplanar then $$[\overline{b}\overline{c}\overline{d}]+[\overline{c}\overline{a}\overline{d}]+[\overline{a}\overline{b}\overline{d}]=$$
    Solution

  • Question 9
    1 / -0
    A point $$O$$ is the centre of a circle circumscribed about a triangle $$ABC$$, then $$\vec{OA}\sin 2A + \vec{OB}\sin 2B + \vec{OC} \sin 2C $$ is equal to
    Solution
    The position vector of the point $$O$$ with respect to itself is
    $$\displaystyle \dfrac{\vec{OA} \sin 2A + \vec{OB} \sin 2B + \vec{OC} \sin 2C}{\sin  2A + \sin  2B + \sin  2C}$$

    $$\displaystyle \Rightarrow \dfrac{\vec{OA} \sin 2A + \vec{OB} \sin 2B + \vec{OC} \sin 2C}{\sin  2A + \sin 2B + \sin  2C} = \vec{0}$$

    or $$\vec{OA} \sin 2A + \vec{OB}\sin 2B + \vec{OC} \sin 2C = \vec{0}$$
  • Question 10
    1 / -0
    If $$\vec {a}=x\hat {i}+12\hat {j}-\hat {k},\vec {b}=2\hat {i}+2x\hat {j}+\hat {k}$$ and $$\vec {c}=\hat {i}+\hat {k}$$ and given that the vectors $$\vec {a},\vec {b},\vec {c}$$ form a right handed system, then the range of $$x$$ is
    Solution
    For Right handed system
    $$[abc]\neq 0$$
    $$[abc]=\begin{vmatrix}x &12  &-1 \\2  &2x  &1 \\1  &0  &1 \end{vmatrix}$$

    $$0\not\equiv 2x^{2}-12+2x$$
    $$\neq x^{2}-6+x$$
    $$0\neq (x+3)(x-2)$$
    $$x=R-[-3,2]$$
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