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Vector Algebra Test - 27

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Vector Algebra Test - 27
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  • Question 1
    1 / -0
    Let $$\vec{A}=2\hat{i}+7\hat{j},\vec{B}=\hat{i}+2\hat{j}+4\hat{k},\ \displaystyle \vec{C}=\dfrac{9\hat{i}+30\hat{j}+4\hat{k}}{5}$$. 
    The ratio in which $$\vec{C}$$ divides $$\vec{AB}$$ internally is?
    Solution
    As shown in the figure let $$\vec{C}$$ divide $$\vec{AB}$$ in a ratio $$K:1$$
    So $$C= \dfrac{KB+A}{K+1}$$
    $$\implies \dfrac{9\hat{i}+30\hat{j}+4\hat{k}}{5} = \dfrac{K(\hat{i}+2\hat{j}+4\hat{k})+(2\hat{i}+7\hat{j})}{K+1}$$
    $$\implies 9K\hat{i}+30K\hat{j}+4K\hat{k}+9\hat{i}+30\hat{j}+4\hat{k} = {5K\hat{i}+10K\hat{j}+20K\hat{k}+10\hat{i}+35\hat{j}}$$
    $$\implies 4K\hat{i}+20K\hat{j}-16K\hat{k}=\hat{i}+5\hat{j}-4\hat{k}$$
    $$\implies 4K(\hat{i}+5\hat{j}-4\hat{k})=\hat{i}+5\hat{j}-4\hat{k}$$
    $$\implies K=\dfrac{1}{4}$$
    Hence, $$\vec{C}$$ divide $$\vec{AB}$$ in a ratio $$1:4$$

  • Question 2
    1 / -0
    If the position vectors of $$A,B,C$$ are $$3\hat{i}+\hat{j}-\hat{k}$$, $$\hat{i}-2\hat{j}$$, $$2\hat{i}-\hat{j}{+}3\hat{k}$$ then the position vector of the centroid of the triangle $$ABC$$ is
    Solution
    Centroid of a triangle is given by
    $$G= \dfrac{\vec{A}+\vec{B}+\vec{C}}{3}$$
    Putting the given values we get
    $$= \dfrac{3\hat{i}+\hat{j}-\hat{k}+\hat{i}-2\hat{j}+2\hat{i}-\hat{j}+3\hat{k}}{3}$$
    $$= 2\hat{i}\ + \dfrac{-2}{3}\hat{j}\ +\dfrac{2}{3}\hat{k}$$
  • Question 3
    1 / -0
    The position vectors of points $$\vec{A},\vec{B},\vec{C}$$ are respectively $$\vec{a},\vec{b},\vec{c}$$. If $$P$$ divides $$\vec{AB}$$ in the ratio $$3:4$$ and $$Q$$ divides $$\vec{BC}$$ in the ratio $$2:1$$ both externally then $$\vec{PQ}$$ is

    Solution
    By section formula
    $$\vec p = \dfrac{3 \vec b - 4 \vec a}{3 - 4} = \dfrac{3 \vec b - 4 \vec a}{-1}$$ and $$\vec q = \dfrac{2 \vec c - \vec b}{2 - 1} = \dfrac{2 \vec c - \vec b}{1}$$
    Thus,
    $$\vec {PQ} = \vec q - \vec p $$
    $$\therefore$$ $$\vec {PQ} = 2 \vec c - \vec b - (- 3 \vec b + 4 \vec a)$$
    $$\therefore$$ $$\vec {PQ} = 2 \vec c - \vec b + 3 \vec b - 4 \vec a$$
    $$\therefore$$ $$\vec {PQ} = 2 \vec c + 2 \vec b - 4 \vec a$$

  • Question 4
    1 / -0
    If $$\vec a, \vec b, \vec c, \vec d$$ are the position vectors of the points $$A, B, C, D$$ respectively such that $$3\vec{a}+5\vec{b}-3\vec{c}-5\vec{d}=0$$ then $$AB$$ intersects $$CD$$ in the ratio
    Solution
    Solving the equation
    $$3 \overline a + 5 \overline b = 3 \overline c + 5 \overline d$$
    $$\therefore$$ $$\dfrac{3 \overline a + 5 \overline b}{3 + 5} = \dfrac{3 \overline c + 5 \overline d}{3 + 5}$$
    We see that there is a point such that it's dividing $$AB$$ as well as $$CD$$ in the above format, thus its equivalent to say that $$AB$$ divides $$CD$$ in the ratio $$5 : 3$$ internally.
  • Question 5
    1 / -0
    If $$(2, -1, 2)$$ is the centroid of tetrahedron $$OABC$$ and $$G_{1}$$ is the centroid of $$\Delta ABC$$ then $$|\overline{OG}_{1}|=$$
    Solution
    Centroid of $$OABC= \dfrac{\vec{a}+\vec{b}+\vec{c}+\vec{o}}{4}$$
    $$4\times (2,-1,2)= \vec{a}+\vec{b}+\vec{c}$$
    Centroid of $$\Delta ABC= \dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$$
    $$G_{1}= \dfrac{4}{3}(2,-1,2)$$
    $$\left | OG_{1} \right |= \dfrac{4}{3}\sqrt{2^{2}+1^{2}+2^{2}}$$
    $$= 4$$
  • Question 6
    1 / -0
    If $$\overline{a}=\overline{i}+\overline{j}+\overline{k},\overline{b}=2\overline{i}-3\overline{j}+\overline{k}$$, then $$\displaystyle \dfrac{\overline{a}\times\overline{b}}{|\overline{a}\times\overline{b}|}+\dfrac{\overline{b}\times\overline{a}}{|\overline{b}\times\overline{a}|}=$$
    Solution
    $$\overline{a} \times \overline{b} = -(\overline{b} \times \overline{a})$$
    Also, $$|\overline{a} \times \overline{b}| = |(\overline{b} \times \overline{a})|$$
    Thus, the denominators become equal and the numerator adds up to $$0$$.
    Hence the answer becomes $$0$$.
  • Question 7
    1 / -0
    The position vectors of $$A, B , C, D$$ are $$\overline{a},\overline{b},\ \overline{c},\overline{d}$$ respectively and $$|\overline{a}-\overline{d}|=|\overline{b}-\overline{d}|=|\overline{c}-\overline{d}|$$, then for the triangle $$ABC$$, $$D$$ is
    Solution
    $$|\overline{a}-\overline{d}|=|\overline{b}-\overline{d}|=|\overline{c}-\overline{d}|$$
    $$\Rightarrow $$ Point $$D$$ is equidistant from the points $$A,B,C$$ or the vertices of triangle.
    $$\Rightarrow  D$$ must be the circumcentre of the triangle.  

  • Question 8
    1 / -0
    Let $$\overrightarrow { b } =4\hat { i } +3\hat { j } $$ and $$\overrightarrow{c}$$ be two vector perpendicular to each other in the $$xy$$-plane. Then a vector in the same plane having projections $$1$$ and $$2$$ along $$\overrightarrow{b}$$ and $$\overrightarrow{c}$$, respectively, is
    Solution
    $$\overrightarrow { c } =x\hat { i } +y\hat { j } $$
    Now, $$\overrightarrow { b } \bot \overrightarrow { c } \Rightarrow \overrightarrow { b } .\overrightarrow { c } =0$$
    $$\displaystyle \Rightarrow 4x+3y=0\Rightarrow \dfrac { x }{ 3 } =\frac { y }{ -4 } =\lambda \\ \Rightarrow x=3\lambda ,y=-4\lambda \\ \therefore \overrightarrow { c } =\lambda \left( 3\hat { i } -4\hat { j }  \right) $$
    Let the required vector be $$\alpha =p\hat { i } +q\hat { j } $$.
    Then the projections of $$\overrightarrow{\alpha}$$ on $$\overrightarrow{b}$$ and $$\overrightarrow{c}$$ are $$\displaystyle \frac { \overrightarrow { \alpha  } .\overrightarrow { b }  }{ \left| b \right|  } $$ and $$\displaystyle \dfrac { \overrightarrow { \alpha  } .\overrightarrow { c }  }{ \left| c \right|  } $$ respectively.
    $$\displaystyle \therefore \dfrac { \overrightarrow { \alpha  } .\overrightarrow { b }  }{ \left| \overrightarrow { b }  \right|  } =1$$ and $$\displaystyle \dfrac { \overrightarrow { \alpha  } .\overrightarrow { c }  }{ \left| \overrightarrow { c }  \right|  } =2$$
    $$\Rightarrow 4p+3q=5$$ and $$3p-4q=10$$
    $$\Rightarrow p=2,q=-1$$
    Hence, the required vector is $$2\hat{i}-\hat{j}$$.
  • Question 9
    1 / -0
    The angle between the Vectors $$\vec{a}\times\vec{b}$$ and $$\vec{b}\times\vec{a}$$ is
    Solution
    $$\vec{b}\times\vec{a}=-\vec{a}\times\vec{b}$$
    $$(\vec{a}\times \vec{b})(\vec{b}\times \vec{a})=|\vec{a}\times \vec{b}||\vec{b}\times \vec{a}|cos \theta$$
    $$\Rightarrow \dfrac{-|\vec{a}\times \vec{b}|}{|\vec{a}\times\vec{b}|}=cos \theta$$
    $$\Rightarrow \theta =\pi$$
  • Question 10
    1 / -0
    The vectors $$\vec{a},\ \vec{b},\ \vec{a}\times \vec{b}$$ form
    Solution
    $$\vec{a}.(\vec{a}\times \vec{b})=0$$
    $$\vec{b}.(\vec{a}\times \vec{b})=0$$
    $$\vec{a}$$ & $$\vec{b} $$ are$$ \perp$$ to $$\vec{a}\times \vec{b}$$
    By Right hand rule, thumb point in direction of $$\vec{a}\times \vec{b}$$ when we curl our finger from $$\vec{a} $$ to $$\vec{b}$$

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