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Vector Algebra Test - 57

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Vector Algebra Test - 57
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  • Question 1
    1 / -0
    Let $$\overrightarrow{a},\overrightarrow{b}$$ be the position vectors of points $$A$$ and $$B$$ with respect to $$O$$ and $$\left|\overrightarrow{a}\right|=a,\left|\overrightarrow{b}\right|=b$$ the points $$C$$ and $$D$$ divides $$A$$ internally and externally in the ratio $$2:3$$ If $$\overrightarrow{OC}$$ and $$\overrightarrow{OD}$$ are perpendicular then
    Solution
    Using section formula,(internally) $$\dfrac{n\overrightarrow{a}+m\overrightarrow{b}}{m+n}$$
    $$\overrightarrow{OC}=\dfrac{3\times\overrightarrow{a}+2\times\overrightarrow{b}}{5}$$
    Using section formula,(externally) $$\dfrac{n\overrightarrow{a}-m\overrightarrow{b}}{n-m}$$
    $$\overrightarrow{OD}=3\times\overrightarrow{a}-2\times\overrightarrow{b}$$
    $$\overrightarrow{OC}\bot$$ to $$\overrightarrow{OD}$$(given)
    $$\Rightarrow\left(\dfrac{3\times\overrightarrow{a}+2\times\overrightarrow{b}}{5}\right)\left(3\times\overrightarrow{a}-2\times\overrightarrow{b}\right)=0$$
    $$\left(3\times\overrightarrow{a}+2\times\overrightarrow{b}\right)\left(3\times\overrightarrow{a}-2\times\overrightarrow{b}\right)=0$$
    $$9\times{a}^{2}=4\times{b}^{2}$$
    or $$9{a}^{2}=4{b}^{2}$$
  • Question 2
    1 / -0
    If $$\vec{a}+\vec{b}+\vec{c}=vec{0}$$ then $$\vec{a}\times \vec{b}=?$$
    Solution
    $$\vec { a } +\vec { b } +\vec { c } =\vec { 0 }\ \dots(1) \, \, \, \, then,\, \,  \\ \vec { a } \times \vec { b } =? \\ take\, \, (1)\, \, cross\, \, product\, \, with\, \, \vec { b } \, \,. \\ \Rightarrow \left( { \vec { a } +\vec { b } +\vec { c }  } \right) \times \vec { b } =\vec { 0 } \times \vec { b }  \\ \Rightarrow \vec { a } \times \vec { b } +\vec { b } \times \vec { b } +\vec { c } \times \vec { b } =0 \\ \to \vec { a } \times \vec { b } +0-\vec { b } \times \vec { c } =0 \\ \therefore \vec { a } \times \vec { b } =\vec { b } \times \vec { c } \, \, \, \, $$
  • Question 3
    1 / -0
    The ratio in which $$i+2j+3k$$ divides the join of $$-2i+3j+5k$$ and $$7i-k$$ is?
    Solution
    $$\begin{array}{l} i+2j+3k=\left( { 1,2,3 } \right)  \\ -2i+3j+5k=\left( { -2,3,5 } \right)  \\ 7i-k=\left( { 7,0,-1 } \right)  \\ 1=\frac { { 7m-2 } }{ { m+1 } }  \\ m+1=7m-2 \\ 3=6m \\ m=\frac { 1 }{ 2 }  \\ So,ratio\, \, is\, \, 1:2 \\ Option\, \, B\, \, \, is\, \, correct\, \, \, answer. \end{array}$$

  • Question 4
    1 / -0
    Let $$\left| \overline { a } +\overline { b }  \right| =\left| \overline { a } -\overline { b }  \right| $$. If $$\left| \overline { a } \times \overline { b }  \right| =\lambda \left| \overline { a }  \right| $$, then $$\lambda=$$
    Solution
    $$|\bar{a}+\bar{b}|=|\bar{a}-\bar{b}|$$
    $$|a|^2+|b|^2+2\bar{a}\cdot \bar{b}=|a|^2+|b|^2-2\bar{a}\cdot\bar{b}$$
    $$4\bar{a}\cdot \bar{b}=0$$
    $$\bar{a}\cdot \bar{b}=0$$
    $$\Rightarrow \bar{a}$$ & $$\bar{b}$$ are perpendicular to each other.
    $$|\bar{a}\times \bar{b}|=|\bar{a}||\bar{b}|\sin\theta$$
    Since $$\bar{a}$$ & $$\bar{b}$$ are perpendicular, $$\theta =90^o$$
    $$=|\bar{a}|\bar{b}|\sin 90^o$$
    $$=|a||b|$$
    $$\therefore \lambda =|b|$$.

  • Question 5
    1 / -0
    Let $$A,B,C$$ be distinct point with position vectors $$\hat{i}+\hat{j}$$, $$\hat{i}-\hat{j}$$, $$p\hat{i}-q\hat{j}+r\hat{k}$$ respectively. Points $$A,B,C$$ are collinear, then which of the following can be correct:
    Solution
    The three points A, B and C are collinear if they satisfy one of the conditions
    $$1) AB+BC=AC, 2). AC+BC=AB $$ and $$ 3). AB+AC=BC$$
    considering the first condition, we have
    $$AB=2$$, $$AC=\sqrt{(1-p)^2+(1+q)^2+r^2}$$ and $$ BC=\sqrt{(1-p)^2+(-1+q)^2+r^2}$$
    To satisfy the condition $$AB+BC=AC$$, $$p, q$$ and $$r$$ take the values $$1, 2$$ and $$0$$ respectively.
    Option D gives the correct answer.
  • Question 6
    1 / -0
    The projection of the vector $$\hat{i}-2\hat{j}+\hat{k}$$ on he vector $$4\hat{i}-4\hat{j}+7\hat{k}$$ is equal to:
    Solution
    The projection of the vector $$\overrightarrow A $$ on the vector $$\overrightarrow B $$ is
    $$\dfrac{{\overrightarrow A  \cdot \overrightarrow B }}{{\left| {\overrightarrow B } \right|}} = \dfrac{{1 \times 4 + \left( { - 2} \right) \times \left( { - 4} \right) + 1 \times 7}}{{\sqrt {{4^2} + {{\left( { - 4} \right)}^2} + {7^2}} }} = \dfrac{{19}}{{\sqrt {81} }} = \dfrac{{19}}{9}$$
  • Question 7
    1 / -0
    If $$\left|\overrightarrow{a} \right|=1$$, the projection of $$\overrightarrow{r}$$ along $$\overrightarrow{a} $$ is $$2$$ and $$\overrightarrow{a}\times \overrightarrow{r}+\overrightarrow{b}=\overrightarrow{r}$$, then $$\overrightarrow{r}=$$
    Solution
    Projection of $$\overrightarrow{r}$$ on $$\overrightarrow{a}$$ is $$2$$ 
    $$\Rightarrow \overrightarrow{r}.\overrightarrow{a}=2$$ -----$$\left(1\right)$$
    $$\overrightarrow{a}\times \overrightarrow{r}+\overrightarrow{b}=\overrightarrow{r}$$ ------$$\left(2\right)$$
    Dot $$\left(1\right)$$ with $$\overrightarrow{a}$$
    $$ \Rightarrow\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{r}.\overrightarrow{a}=2$$ by $$\left(1\right)$$
    cross $$\left(1\right)$$ with $$\overrightarrow{a}$$
    $$\Rightarrow \overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{r}\right)+\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{r}$$
    $$\Rightarrow 0+\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{r}$$
    $$\therefore 2\overrightarrow{r}-\overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{b}=2\overrightarrow{r}$$
    $$\Rightarrow \overrightarrow{r}=\dfrac{1}{2}\left[2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{b}\right]$$
  • Question 8
    1 / -0
    If $$\overrightarrow{u}, \overrightarrow{v}$$ and $$\overrightarrow{w}$$ are three non-coplanar vectors, then $$(\overrightarrow{u} + \overrightarrow{v} - \overrightarrow{w}). (\overrightarrow{u} - \overrightarrow{v}) \times (\overrightarrow{v} - \overrightarrow{w})$$ equals
    Solution

    Using the distributive law, together with the identities

    $$\begin{array}{l} \overrightarrow { a } \times \overrightarrow { a }  \\ \overrightarrow { a } \cdot \left( { \overrightarrow { a } \times \overrightarrow { b }  } \right) =0 \\ \overrightarrow { b } \cdot \left( { \overrightarrow { a } \times \overrightarrow { b }  } \right) =0 \\ We\, get \\ \left( { \overrightarrow { u } +\overrightarrow { v } -\overrightarrow { w }  } \right) \cdot \left( { \left( { \overrightarrow { u } -\overrightarrow { v }  } \right) \times \left( { \overrightarrow { v } -\overrightarrow { w }  } \right)  } \right)  \\ \left( { \overrightarrow { u } \cdot \left( { \left( { \overrightarrow { u } -\overrightarrow { v }  } \right) \times \left( { \overrightarrow { v } -\overrightarrow { w }  } \right)  } \right)  } \right) +\left( { \left( { \overrightarrow { v } -\overrightarrow { w }  } \right) \cdot \left( { \left( { \overrightarrow { u } -\overrightarrow { v }  } \right) \times \left( { \overrightarrow { v } -\overrightarrow { w }  } \right)  } \right)  } \right)  \\ \overrightarrow { u } \cdot \left( { \left( { \overrightarrow { u } -\overrightarrow { v }  } \right) \times \left( { \overrightarrow { v } -\overrightarrow { w }  } \right)  } \right)  \\ \overrightarrow { u } \cdot \left( { \left( { \overrightarrow { u } \times \overrightarrow { v }  } \right) -\left( { \overrightarrow { u } \times \overrightarrow { w }  } \right) +\left( { \overrightarrow { v } \times \overrightarrow { w }  } \right)  } \right)  \\ \overrightarrow { u } \cdot \left( { \overrightarrow { u } \times \overrightarrow { v }  } \right) -\overrightarrow { u } \cdot \left( { \overrightarrow { u } \times \overrightarrow { w }  } \right) +\overrightarrow { u } \cdot \left( { \overrightarrow { v } \times \overrightarrow { w }  } \right)  \\ \overrightarrow { u } \cdot \left( { \overrightarrow { v } \times \overrightarrow { w }  } \right)  \end{array}$$

  • Question 9
    1 / -0
    If $$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}\neq 0$$ then $$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=$$
  • Question 10
    1 / -0

    A unit vector perpendicular to the plane of the triangle ABC with the position vectors  $$\vec a\,\,\,\,\vec b\,\,\vec c$$ of the vectors A,B,C, is 

    Solution

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