Vector Algebra Test

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Vector Algebra Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

If $$C$$ is the mid point of $$AB$$ and $$P$$ is any point out side $$AB$$, then

A
$$\vec{PA}+\vec{PB}+2\vec{PC}=\vec{0}$$

B
$$\vec{PA}+\vec{PB}+\vec{PC}=\vec{0}$$

C
$$\vec{PA}+\vec{PB}=2\vec{PC}$$

D
$$\vec{PA}+\vec{PB}=\vec{PC}$$

Solution

From above figure
$$\vec{PA}=\vec{CA}+\vec{BC}+\vec{PB}$$
$$=2\vec{BC}+\vec{PB}$$
$$=2(\vec{PC}-\vec{PB})+\vec{PB}$$
$$\vec{PA}+\vec{PB}=2\vec{PC}$$
Question 2
1 / -0

$$ABC$$ is a triangle and $$P$$ is any point on $$BC$$. If $$PQ$$ is the resultant of the vectors $$\vec {AP},\ \vec {PB}$$ and $$\vec{PC}$$ then $$ACQB$$ is

A
Rectangle

B
Square

C
Rhombus

D
Parallelogram

Solution

$$\vec{PQ}=\vec{AP}+\vec{PB}+\vec{PC}$$
$$\vec{PQ}-\vec{PB}=\vec{AP}+\vec{PC}$$
$$\vec{BQ}=\vec{AC}$$ --------(1)
$$\vec{PQ}-\vec{PC}=\vec{AP}+\vec{PB}$$
$$\vec{CQ}=\vec{AB}$$ ---------(2)
So $$ACQB $$ is Parallelogram.
Question 3
1 / -0

If $$\vec{b}$$ is the vector whose initial point divides the joining $$5\hat{i}$$ and $$5\hat{j}$$ in the ratio $$\lambda :$$ $$1$$ and terminal point is at origin. lf $$|\vec{b}|\leq\sqrt{37}$$, then $$\lambda\in$$.

A
$$(-\displaystyle \infty, -6]\cup \left [-\dfrac{1}{6}, \infty \right)$$

B
$$(-\displaystyle \infty, -3)\cup \left [-\dfrac{1}{4}, \infty \right)$$

C
$$(-\infty, 0)\cup \left(\dfrac{1}{2} , \infty\right)$$

D
$$\left [-6, -\displaystyle \dfrac{1}{6}\right]$$

Solution

Apply section formula and we get,
$$\vec b = \dfrac{5 \hat i + 5 \lambda \hat j}{\lambda + 1}$$
$$|\vec b| \leq \sqrt{37}$$
$$\therefore$$  $$25 \times {\lambda}^2 + 25 \leq ({\lambda + 1}^2) \times 37$$
$$\therefore$$  $$25 \times {\lambda}^2 + 25 \leq 37 \times ({\lambda}^2 + 2\lambda + 1)$$
$$\therefore$$  $$25 \times {\lambda}^2 + 25 \leq 37\times {\lambda}^2 + 74 \lambda + 37$$
$$\therefore 12 \times {\lambda}^2 + 74 \lambda + 12 \geq 0.$$
$$\therefore 6 \times {\lambda}^2 + 37 \lambda + 6 \geq 0$$
$$\therefore \lambda \in ( - \infty , -6] \cup \left [ \dfrac{-1}{6}, \infty \right)$$
Question 4
1 / -0

lf the Vector $$\overline{\mathrm{c}},\ \vec{\mathrm{a}}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}},\ \vec{\mathrm{b}}=\hat{\mathrm{j}}$$ are such that $$\vec{\mathrm{a}},\ \vec{\mathrm{c}},\ \vec{\mathrm{b}}$$ form $$\mathrm{R}.\mathrm{H}.\ \mathrm{S}$$ then $$\vec{\mathrm{c}}=$$

A
$$\mathrm{z}\hat{\mathrm{i}}-\mathrm{x}\hat{\mathrm{k}}$$

B
$$\mathrm{x}\hat{\mathrm{i}}-\mathrm{z}\hat{\mathrm{k}}$$

C
$$\mathrm{x}\hat{\mathrm{j}}-\mathrm{y}\hat{\mathrm{k}}$$

D
$$\mathrm{y}\hat{\mathrm{j}}-\mathrm{x}\hat{\mathrm{k}}$$

Solution

$$\vec{b}\times \vec{a}=\vec{c} (As \vec{a}, \vec{c}, \vec{b} are in R.H.S)$$

$$\vec{b} \times \vec{a}=\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\\
0 & 1 & 0\\
x & y & z
\end{vmatrix} =z\hat{i}-x\hat{k}$$

So $$\vec{c}=z\vec{i}-x\vec{k}$$
Question 5
1 / -0

If $$\vec{r}=3\hat{i}+2\hat{j}-5\hat{k},\vec{a}=2\hat{i}-\hat{j}+\hat{k}$$, $$\vec{b}=\hat{i}+3\hat{j}-2\hat{k},\ \vec{c}=-2\hat{i}+\hat{j}-3\hat{k}$$ such that $$\vec{r}=\lambda\vec{a}+\mu\vec{b}+v\vec{c}$$, then $$\mu,\ \displaystyle \frac{\lambda}{2}$$ , $$v$$ are in

A
$$H.P$$

B
$$G.P$$

C
$$A.G.P$$

D
$$A.P$$

Solution

$$(3,2,-5)=\lambda (2,-1,1)+\mu (1,3,-2)+\gamma (-2,1,-3)$$
$$3=2\lambda +\mu -2\gamma $$
$$2=-\lambda +3\mu +\gamma $$
$$-5=\lambda -2\mu -3\gamma $$
$$\begin{bmatrix}
2 & 1 & -2\\
-1 & 3 & 1\\
1 & 2 & -3
\end{bmatrix}
\begin{bmatrix}
\lambda\\
\mu\\
\nu
\end{bmatrix}=\begin{bmatrix}
3\\
2\\
-5
\end{bmatrix}$$
$$R_{2}\rightarrow R_{2}+R_{3}$$
$$\begin{bmatrix}
2 & 1 & -2\\
0 & 1 & -2\\
1 & -2 & -3
\end{bmatrix}
\begin{bmatrix}
\lambda\\
\mu\\
\nu
\end{bmatrix}=\begin{bmatrix}
3\\
-3\\
-5
\end{bmatrix}$$
$$R_{1}\rightarrow R_{1}-R_{2}$$
$$\begin{bmatrix}
2 & 0 & 0\\
0 & 1 & -2\\
1 & -2 & -3
\end{bmatrix}
\begin{bmatrix}
\lambda\\
\mu\\
\nu
\end{bmatrix}=\begin{bmatrix}
6\\
-3\\
-5
\end{bmatrix}$$
$$\lambda =3 \gamma =2 \mu =1$$
Question 6
1 / -0

$$\overline{a}=x\hat {i}+y\hat {j}+z\hat {k},\ \overline{b}=\hat {j}$$, then the vector $$\overline{c}$$ for which $$\overline{a},\overline{b},\ \overline{c}$$ form a right hand triangle

A
$$x(\hat {i}-\hat {k})$$

B
$$\hat {0}$$

C
$$-z\hat {i}+x\hat {k}$$

D
$$y\hat {j}$$

Solution

$$\overline{a},\overline{b},\overline{c}$$ form a right handed triangle.
When this holds true, $$\overline{c} = \overline{a} \times \overline{b}$$
Thus, taking cross product , we get $$\overline{c} = -z \overline{i} + x \overline{k}$$
Question 7
1 / -0

If $$\vec a=\hat {i}+2\hat {j}-3\hat {k}$$ and $$\vec {b}=2\hat {i}-\hat {j}-\hat {k}$$ then the ratio between the projection of $$\vec b$$ on $$\vec {a}$$ and the projection of $$\vec {a}$$ on $$\vec {b}$$ is

A
$$\sqrt{5}:\sqrt{7}$$

B
$$\sqrt{3}:\sqrt{7}$$

C
$$\sqrt{7}:\sqrt{3}$$

D
$$\sqrt{7}:\sqrt{5}$$

Solution

Projection of $$\overline{b}$$ on $$\overline{a}=\dfrac{\overline{b}\cdot\overline{a}}{|\overline{a}|}$$
Projection of $$\overline{a}$$ on $$\overline{b}=\dfrac{\overline{b}\cdot\overline{a}}{|\overline{b}|}$$
Ratios$$=|\overline{b}|:|\overline{a}|$$
$$=\sqrt{6} :\sqrt{14} = \sqrt{3} : \sqrt{7}$$
Question 8
1 / -0

Which of the following is a true statement.

A
$$\vec {a}\times\vec {b})\times\vec {c}$$ is coplanar with $$\vec {c}$$

B
$$(\vec {a}\times\vec {b})\times\vec {c}$$ is perpendicular to $$\vec {a}$$

C
$$(\vec {a}\times\vec {b})\times\vec {c}$$ is perpendicular to $$\vec {b}$$

D
$$(\vec {a}\times\vec {b})\times\vec {c}$$ is perpendicular to $$\vec {c}$$

Solution

We know that $$\vec {a}\times\vec {b}$$ is perpendicular to both $$\vec {a}$$ and $$\vec {b}$$
Hence $$(\vec {a}\times \vec {b})\times \vec {c}$$ is perpendicular to both $$\vec {a}\times \vec {b}$$ and $$\vec {c}$$
Question 9
1 / -0

Given, $$|\vec {a}|=|\vec {b}|=1$$ and $$|\vec {a}+\vec {b}|=\sqrt{3}$$. If $$\vec {c}$$ be a vector such that $$\vec {c}-\vec {a}-2\vec {b}=3(\vec {a}\times\vec {b})$$ , then $$\vec {c}.\vec {b}$$ is equal to

A
$$-\displaystyle \dfrac{1}{2}$$

B
$$\displaystyle \dfrac{1}{2}$$

C
$$\displaystyle \dfrac{3}{2}$$

D
$$\displaystyle \dfrac{5}{2}$$

Solution

We have $$\overline{c}-\overline{b}=\overline{a}-\overline{b}+2$$
$$\left | \overline{a}+\overline{b} \right |=\sqrt{\bar a^{2}+\bar b^{2}+2\bar \cdot \bar b}$$$$3=1+1+2\left ( \overline{a\cdot} \overline{b}\right )$$
$$\dfrac {1}{2}=\overline{a}-\overline{b}$$
$$\overline{c}-\overline b=\dfrac {5}{2}$$
Question 10
1 / -0

If the position vector of a point $$A$$ is $$\overrightarrow { a } +\overrightarrow { 2b } $$ and $$\overrightarrow { a }$$ divides $$\overrightarrow{AB}$$ in the ratio $$2:3$$, then the position vector of $$B$$ is

A
$$\overrightarrow { 2a } -\overrightarrow { b } $$

B
$$\overrightarrow { b } -\overrightarrow { 2a } $$

C
$$\overrightarrow { a } -\overrightarrow { 3b } $$

D
$$\overrightarrow { b } $$

Solution

If $$\overrightarrow{x}$$ is the position vector of $$B$$, then $$\overrightarrow{a}$$ divides $$\overrightarrow{AB}$$ in the ration $$42:3$$
$$\displaystyle \therefore \overrightarrow { a } =\dfrac { 2\overrightarrow { x } +3\left( \overrightarrow { a } +2\overrightarrow { b } \right) }{ 2+3 } $$

$$\Rightarrow 5\overrightarrow { a } -3\overrightarrow { a } -6\overrightarrow { b } =2\overrightarrow { x } \Rightarrow \overrightarrow { x } =\overrightarrow { a } -3\overrightarrow { b } $$

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Vector Algebra Test - 29

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Vector Algebra Test - 29

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SHARING IS CARING

Question 1

$$\vec{PA}+\vec{PB}+2\vec{PC}=\vec{0}$$

$$\vec{PA}+\vec{PB}+\vec{PC}=\vec{0}$$

$$\vec{PA}+\vec{PB}=2\vec{PC}$$

$$\vec{PA}+\vec{PB}=\vec{PC}$$

Solution

Question 2

Rectangle

Square

Rhombus

Parallelogram

Solution

Question 3

$$(-\displaystyle \infty, -6]\cup \left [-\dfrac{1}{6}, \infty \right)$$

$$(-\displaystyle \infty, -3)\cup \left [-\dfrac{1}{4}, \infty \right)$$

$$(-\infty, 0)\cup \left(\dfrac{1}{2} , \infty\right)$$

$$\left [-6, -\displaystyle \dfrac{1}{6}\right]$$

Solution

Question 4

$$\mathrm{z}\hat{\mathrm{i}}-\mathrm{x}\hat{\mathrm{k}}$$

$$\mathrm{x}\hat{\mathrm{i}}-\mathrm{z}\hat{\mathrm{k}}$$

$$\mathrm{x}\hat{\mathrm{j}}-\mathrm{y}\hat{\mathrm{k}}$$

$$\mathrm{y}\hat{\mathrm{j}}-\mathrm{x}\hat{\mathrm{k}}$$

Solution

Question 5

$$H.P$$

$$G.P$$

$$A.G.P$$

$$A.P$$

Solution

Question 6

$$x(\hat {i}-\hat {k})$$

$$\hat {0}$$

$$-z\hat {i}+x\hat {k}$$

$$y\hat {j}$$

Solution

Question 7

$$\sqrt{5}:\sqrt{7}$$

$$\sqrt{3}:\sqrt{7}$$

$$\sqrt{7}:\sqrt{3}$$

$$\sqrt{7}:\sqrt{5}$$

Solution

Question 8

$$\vec {a}\times\vec {b})\times\vec {c}$$ is coplanar with $$\vec {c}$$

$$(\vec {a}\times\vec {b})\times\vec {c}$$ is perpendicular to $$\vec {a}$$

$$(\vec {a}\times\vec {b})\times\vec {c}$$ is perpendicular to $$\vec {b}$$

$$(\vec {a}\times\vec {b})\times\vec {c}$$ is perpendicular to $$\vec {c}$$

Solution

Question 9

$$-\displaystyle \dfrac{1}{2}$$

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

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

$$\displaystyle \dfrac{5}{2}$$

Solution

Question 10

$$\overrightarrow { 2a } -\overrightarrow { b } $$

$$\overrightarrow { b } -\overrightarrow { 2a } $$

$$\overrightarrow { a } -\overrightarrow { 3b } $$

$$\overrightarrow { b } $$

Solution

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