MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 12

'$$P$$' is a point inside the triangle $$ABC$$, such that $$\displaystyle BC\left ( \vec{PA} \right )+CA\left ( \vec{PB} \right )+AB\left ( \vec{PC} \right )=0,$$ then for the triangle $$ABC$$ the point $$P$$ is its :

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Incentre
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Circumcentre
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Centroid
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Orthocentre

Let the pairs $$\vec{a}, \vec{b}$$ and $$\vec{c}, \vec{d}$$ each determine a plane, then the planes are parallel if

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$$(\vec{a} \times \vec{c}) \times (\vec{b} \times \vec{d}) = \vec{0}$$
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$$(\vec{a} \times \vec{c}) \cdot (\vec{b} \times \vec{d}) = \vec{0}$$
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$$(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$$
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$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = \vec{0}$$

Let $$\vec{a},\vec{b},\vec{c}$$ be vectors of length $$3,4,5$$ respectively. Let $$\vec{a}$$ be perpendicular to $$\vec{b}+\vec{c},\vec{b}\,to\,\vec{c}+\vec{a}$$ and $$\vec{c}\,to\,\vec{a}+\vec{b}$$. Then $$\begin{vmatrix}\vec{a}+\vec{b}+\vec{c}\end{vmatrix}$$ is

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$$\;2\sqrt{5}$$
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$$\;2\sqrt{2}$$
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$$\;10\sqrt{5}$$
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$$\;5\sqrt{2}$$

$$ABCDEF$$ is a regular hexagon . The centre of hexagon is a point O. Then the value of
$$\overrightarrow{AB}+ \overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}$$ is

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$$2\overrightarrow{AO}$$
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$$4\overrightarrow{AO}$$
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$$6\overrightarrow{AO}$$
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Zero

The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors a, b, c such that a.b = b.c = c.a = 1/What is the volume of the parallelopipe.

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$$\dfrac{1}{\sqrt{2}}$$
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$$\dfrac{1}{\sqrt{3}}$$
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$$\dfrac{3}{\sqrt{2}}$$
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None of these

If the sum of two unit vectors is also a unit vector, then the angle between the two vectors is

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$$\dfrac{\pi}{3}$$
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$$\dfrac{2\pi}{3}$$
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$$\dfrac{\pi}{4}$$
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None of these

Let $$b = 4i + 3j $$ and $$c$$ be two vectors perpendicular to each other in the xy-plane. If $$ r_i, \ i =1, 2 ... n$$, are the vectors in the same plane having projections $$1 $$ and $$2$$ along $$b$$ and $$c$$ respectively then $$ \displaystyle \sum_{i=1}^{n} \left| r_{i}\right|^{2}$$ is equal to

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$$20$$
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$$10$$
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$$4$$
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$$7$$

The magnitude of the projection of the vector $$\overline{a} =4\overline{i}-3\overline{j}+2\overline{k}$$ on the line which makes equal angles with the coordinate axes is

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$$\sqrt{2}$$
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$$\sqrt{3}$$
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$$\dfrac{1}{\sqrt{3}}$$
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$$\dfrac{1}{\sqrt{2}}$$

Let $$ x_{0}$$ and $$x_{1}$$ be the critical points of $$\displaystyle f(x) =\int_{1}^{x}(t(t + 1) (t + 2) (t + 3)- 24).dt $$ and $$ \vec r$$ & $$\vec r'$$ be the parallel vectors with $$\left| \vec r \right| =\left| x_{0}\right| $$ and $$\left| \vec r\ ' \right| =\left|x_{1}\right| ,$$ then $$ \vec r\cdot \vec r\ '$$ is equal to

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$$24$$
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$$16$$
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$$8$$
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$$4$$

If one point on the vector $$2i -4j-k$$ is $$(2,1,3)$$,the other point is?

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$$(-4,3,2)$$
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$$(4,-3,-2)$$
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$$(3,2,1)$$
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$$(4,-3,2)$$

If the vectors $$\overline { c } ,\overline { a } =x\hat { i } +y\hat { j } +z\hat { k } $$, and $$\overline { b } =\hat { j }$$ are such that $$\overline { a } ,\overline { c } \ and \ \overline { b }$$ from a right handed system, then $$\overline { c }$$ is

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$$z\hat { i-x } \hat { k }$$
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$$\overline { 0 }$$
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$$y\hat { j }$$
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$$-z\hat { i+x } \hat { k }$$

A point $$C = \dfrac {5\overline {a} + 4\overline {b} - 5\overline {c}}{3}$$ divides the line joining the points $$A$$ and $$B = 2\overline {a} + 3\overline {b} - 4\overline {c}$$ in the ratio $$2 : 1$$, then the position vector of $$A$$ is

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$$\overline {a} + 3\overline {b} - 4\overline {c}$$
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$$2\overline {a} - 3\overline {b} + 4\overline {c}$$
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$$2\overline {a} + 3\overline {b} + 4\overline {c}$$
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$$\overline {a} - 2\overline {b} + 3\overline {c}$$

If $$\overrightarrow{a} = 2\hat{i} - \hat{j} + \hat{k}$$, $$\overrightarrow{b} = \hat{i} + 2\hat{j} - \hat{k}$$, $$\overrightarrow{c} = \hat{i} + \hat{j} - 2\hat{k}$$, then a vector in the plane of $$\hat{b}$$ and $$\hat{c}$$ whose projection on $$\hat{a}$$ is a magnitude of $$\sqrt{\frac{2}{3}}$$ is

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$$2\hat{i} + 3\hat{j} - 3\hat{k}$$
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$$2\hat{i} + 3\hat{j} + 3\hat{k}$$
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$$-2\hat{i} - \hat{j} + 5\hat{k}$$
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$$2\hat{i} + \hat{j} + 5\hat{k}$$

If P(x, y, z) is a point on the line segment joining Q(2, 2, 4) and R(3, 5, 6) such that the projections of $$\overrightarrow{OP}$$ on the axes are $$\dfrac{13}{5}, \dfrac{19}{5}, \dfrac{26}{5}$$ respectively, then P divides QR in the ratio

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1 : 2
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3 : 2
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2 : 3
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1 : 3

What are coinitial vectors.?

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Two or more vectors having the same magnitude are called coinitial vectors.
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Two or more vectors are said to be coinitial if they are parallel to the same line, irrespective of their magnitudes and directions.
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Two or more vectors having the same initial point are called coinitial vectors.
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None of the above

If three non-zero vectors are $$a=a_1i+a_2j+a_3k , \, b=b_1i+b_2j+b_3k$$ and $$ c=c_1i+c_2j+c_3k$$ . If c is the unit vector perpendicular to the vectors a and b the angle between a and b is $$\dfrac{\pi}{6}$$ , then $$\begin{vmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}$$ is equal to

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0
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$$ \dfrac{3(\sum a^2_1)(\sum b^2_1) \sum c^2_1}{4}$$
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1
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$$\dfrac{(\sum a^2_1)(\sum b^2_1)}{4}$$

A unit vector a makes an angel $$\Pi /4$$ with the z-axis. If a+i+j is a unit vector, then a can be equal to

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$$\dfrac { i }{ 2 } -\dfrac { j }{ 2 } +\dfrac { k }{ \sqrt { 2 } } $$
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$$\dfrac { i }{ 2 } -\dfrac { j }{ 2 } -\dfrac { k }{ \sqrt { 2 } } $$
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$$-\dfrac { i }{ 2 } -\dfrac { j }{ 2 } +\dfrac { k }{ \sqrt { 2 } } $$
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None

Let $$\vec{u},\ \vec{v},\ \vec{w}$$ be such that $$\left| \vec { u } \right| =1$$, $$\left| \vec { v } \right| =2$$, $$\left| \vec { w } \right| =3$$. If the projection of $$\vec{v}$$ along $$\vec{u}$$ is equal to projection of $$\vec{w}$$ along $$\vec{u}$$ and $$\vec{v}$$ and $$\vec{w}$$ are perpendicular to each other then $$\left| \bar { u } -\bar { v } +\bar { w } \right|$$ equals-

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$$2$$
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$$\sqrt{7}$$
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$$\sqrt{14}$$
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$$14$$

Let $$\vec { p } $$ and $$\vec { q } $$ be the position vectors of the points $$P$$ and $$Q$$ respectively with respect to origin $$O$$. The points $$R$$ and $$S$$ divide $$PQ$$ internally and externally respectively in the ratio $$2:3$$. If $$\overrightarrow { OR } $$ and $$\overrightarrow { OS } $$ are perpendicular, then which one of the following is correct?

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$$9{ p }^{ 2 }=4{ q }^{ 2 }$$
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$$4{ p }^{ 2 }=4{ 9q }^{ 2 }$$
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$$9p=4q$$
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$$4p=9q$$

Let $$O$$ be an interior point of $$\triangle ABC$$ such that $$\overline {OA} + 2\overline {AB} + 3\overline {OC} = \overline {O}$$, then the ratio of $$\triangle ABC$$ to area of $$\triangle AOC$$ is

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$$2$$
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$$\dfrac {3}{2}$$
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$$3$$
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$$\dfrac {5}{2}$$

Explanation

Refer to the figure. Let O be the origin.

Let $$\bar { OA } =\bar { a } $$

$$\bar { OB } =\bar { b } $$

$$\bar { OC } =\bar { c } $$

Given, $$\bar { OA } +2\bar { OB } +3\bar { OC } =0$$

$$\therefore \bar { a } +2\bar { b } +3\bar { c } =0$$

$$\therefore \bar { a } =-2\bar { b } -3\bar { c } $$ (1)

1) Area of triangle ABC is given by,

$$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \bar { AB } \times \bar { BC } \right| $$

$$\therefore A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( \bar { b } -\bar { a } \right) \times \left( \bar { c } -\bar { b } \right) \right| $$

From equation (1),

$$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( \bar { b } -\left( -2\bar { b } -3\bar { c } \right) \right) \times \left( \bar { c } -\bar { b } \right) \right| $$

$$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( \bar { b } +2\bar { b } +3\bar { c } \right) \times \left( \bar { c } -\bar { b } \right) \right| $$

$$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \left| \left( 3\bar { b } +3\bar { c } \right) \times \left( \bar { c } -\bar { b } \right) \right| $$

$$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } +\bar { c } \right) \times \left( \bar { c } -\bar { b } \right) \right| $$

$$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) -\left( \bar { b } \times \bar { b } \right) +\left( \bar { c } \times \bar { c } \right) -\left( \bar { c } \times \bar { b } \right) \right| $$

But, $$\bar { b } \times \bar { b } =0$$ and $$\bar { c } \times \bar { c } =0$$

$$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) -\left( \bar { c } \times \bar { b } \right) \right| $$

$$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) +\left( \bar { b } \times \bar { c } \right) \right| $$

$$\left( \because \left( \bar { a } \times \bar { b } \right) =-\left( \bar { b } \times \bar { a } \right) \right) $$

$$\therefore A\left( \triangle ABC \right) =\frac { 3 }{ 2 } \left| 2\left( \bar { b } \times \bar { c } \right) \right| $$

$$\therefore A\left( \triangle ABC \right) =3\left| \left( \bar { b } \times \bar { c } \right) \right| $$ (2)

2) Area of triangle AOC is given by,

$$\therefore \left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| \bar { OA } \times \bar { OC } \right| $$

$$\therefore \left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| \bar { a } \times \bar { c } \right| $$

From equation (1),

$$A\left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| \left( -2\bar { b } -3\bar { c } \right) \times \bar { c } \right| $$

$$\therefore A\left( \triangle AOC \right) =\frac { 1 }{ 2 } \left| -2\left( \bar { b } \times \bar { c } \right) -3\left( \bar { c } \times \bar { c } \right) \right| $$

$$\therefore A\left( \triangle AOC \right) =\frac { 2 }{ 2 } \left| \left( \bar { b } \times \bar { c } \right) \right| $$

$$\therefore A\left( \triangle AOC \right) =\left| \left( \bar { b } \times \bar { c } \right) \right| $$ (3)

Now, $$\frac { A\left( \triangle ABC \right) }{ A\left( \triangle AOC \right) } =\frac { 3\left| \left( \bar { b } \times \bar { c } \right) \right| }{ \left| \left( \bar { b } \times \bar { c } \right) \right| } $$

$$\frac { A\left( \triangle ABC \right) }{ A\left( \triangle AOC \right) } =3$$

The set of values of $$'c'$$ for which the angle between the vectors $$(cx\hat{i}-6\hat{j}+3\hat{k})$$ and $$(x\hat{i}-2\hat{j}+2cx\hat{k})$$ is acute for every $$x\in R$$ is

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$$(0,\ 4/3]$$
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$$[0,\ 4/3)$$
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$$(11/9,\ 4/3)$$
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$$[0,\ 4/3]$$

If $$\vec { a } ,\vec { b }\ and\ \vec { c }$$ unit vector satisfying $$\left| \vec { a } -\vec { b } \right| +\left| \vec { b } -\vec { c } \right| +\left| \vec { c } -\vec { a } \right| =9$$, then $$\left| 2\vec { a } +7\vec { b } +7\vec { c } \right|$$ is equal to

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2
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3
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4
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5

Let $$\vec {a}, \vec {b}$$ and $$\vec {c}$$ be vectors forming right hand triad. Let
$$\vec {p} = \dfrac {\vec {b} \times \vec {c}}{[\vec {a} \vec {b} \vec {c}]}, \vec {q} = \dfrac {\vec {c} \times \vec {c}}{[\vec {a} \vec {b} \vec {c}]}$$ and $$\vec {r} = \dfrac {\vec {a}\times \vec {b}}{[\vec {a}\vec {b}\vec {c}]}$$ if $$x\epsilon R^{+}$$ then

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$$x[\vec {a}\vec {b}\vec {c}] + \dfrac {[\vec {p}\vec {q}\vec {r}]}{x}$$ has least value $$2$$
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$$x^{4} [\vec {a}\vec {b}\vec {c}]^{2} + \dfrac {[\vec {p}\vec {q}\vec {r}]}{x^{2}}$$ has least value $$(3/2^{2/3})$$
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$$[\vec {p}\vec {q}\vec {r}] > 0$$
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$$x[\vec {p}\vec {q}\vec {r}]$$ has maximum value $$7$$

A vector $$\vec{a}$$ has components $$2$$ p and $$1$$ with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to the new system, $$\vec{a}$$ components $$p+1$$ and $$1$$, then?

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$$p=0$$
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$$p=1$$ or $$p=-\dfrac{1}{3}$$
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$$p=-1$$ or $$p=\dfrac{1}{3}$$
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$$p=1$$ or $$p=-1$$

The vector $$(\hat {i}\times \vec {a}.\vec {b})\hat {i} + (\hat {j} \times \vec {a}.\vec {b})\hat {j} + (\hat {k} \times \vec {a} . \vec {b})\hat {k}$$ is equal to

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$$\vec {b}\times \vec {a}$$
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$$\vec {a}$$
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$$\vec {a}\times \vec {b}$$
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$$\vec {b}$$

If $$ABCDE$$ is a pentagon, then $$\vec {AB}+\vec {AE}+\vec {BC}+\vec {DC}+\vec {ED}+\vec {AC}$$ equals

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$$3 \vec {AD}$$
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$$3 \vec {AC}$$
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$$3 \vec {BE}$$
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$$3 \vec {CE}$$

The x-y plane divides the line joining the points $$(-1, 3, 4)$$ and $$(2, -5, 6)$$:

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internally in the ratio $$2:3$$
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externally in the ratio $$2:3$$
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internally in the ratio $$3:2$$
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externally in the ratio $$3:2$$

If $$\vec {a},\vec {b}$$ and $$\vec {c}$$ are those mutually perpendicular vectors, then the projection of the vector $$\left( l \dfrac{\bar {a}}{|\bar {a}|}+m\dfrac{\bar {b}}{|\bar {b}|}+n\dfrac{(\bar {a}\times \bar {b})}{|\bar {a} \times \bar {b}|}\right)$$ along bisector of vectors $$\vec {a}$$ and $$\vec {a}$$ may be given as ?

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$$\dfrac{l^{2}+m^{2}}{\sqrt{l^{2}+m^{2}+n^{2}}}$$
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$$\sqrt{l^{2}+m^{2}+n^{2}}$$
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$$\dfrac{\sqrt{l^{2}+m^{2}}}{\sqrt{l^{2}+m^{2}+n^{2}}}$$
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$$\dfrac{l+m}{\sqrt{2}}$$

Let $$\overline {a}, \overline {b}$$ be two noncollinear vectors. If $$\overline {OA}=(x+4y)\overline {a}+(2x+y+1)\overline {b}, \overline {OB}=(y-2x+2)\overline {a}+(2x-3y-1)\overline {b}$$ and $$3\overline {OA}=2\overline{OB}$$, then $$(x,y)=$$

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$$(1,2)$$
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$$(1,-2)$$
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$$(2,-1)$$
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$$(-2,-1)$$

The projection of $$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$$ on the vector $$\vec{b}=\hat{i}+2\hat{j}-\hat{k}$$ is?

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$$\sqrt{\dfrac{2}{3}}$$
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$$\dfrac{2}{\sqrt{21}}$$
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$$\sqrt{\dfrac{3}{2}}$$
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$$\dfrac{5}{\sqrt{21}}$$

Let $$\hat{a}, \hat{b}$$ and $$\hat{c}$$ be three unit vectors such that $$\hat{a}=\hat{b}+(\hat{b}\times \hat{c})$$, then the possible value(s) of $${|\hat{a}+\hat{b}+\hat{c}|}^{2}$$ can be:

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$$1$$
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$$4$$
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$$16$$
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$$9$$

If $$D\vec A = \vec a,$$ $$A\vec B = \vec b$$ and $$C\vec B = k\vec a$$ where $$k < 0$$ and X, Y are the mid-points of DB & DC respectively, such that $$\left| {\vec a} \right| = 17\& \left| {X\vec Y} \right| = 4$$, then k equal to -

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$${8 \over {17}}$$
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$${{13} \over {17}}$$
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$${{25} \over {17}}$$
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$${4 \over {17}}$$

$$O$$ is the origin in the Cartesian plane. From the origin $$O$$ take point $$A$$ in the North-east direction such that $$|\overline {OA}|=5,B$$ is a point in the North-west direction such that $$|\overline {OB}|=5$$.
Then $$|\overline {OA}-\overline {OB}|$$ is.

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$$25$$
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$$5\sqrt2$$
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$$10\sqrt5$$
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$$\sqrt5$$

The angles of a triangles whose two sides are represented by vectors $$\sqrt(\vec { a } \times \vec { b })$$ and $$\vec{b}-(\vec { a } \vec { b })\vec{a}$$ are in the ratio

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$$1:2:3$$
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$$1:1:2$$
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$$1:3:5$$
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$$1:2:7$$

The vectors $$\vec{u}=\begin{bmatrix} 6 \\ -3 \\ 2 \end{bmatrix}; \vec{v}=\begin{bmatrix} 2 \\ 6 \\ 3 \end{bmatrix}; \vec{w}=\begin{bmatrix} 3 \\ 2 \\ -6 \end{bmatrix}$$

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form a left handed system
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form a right handed system
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are linearly independent
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are such that each is perpendicular to the plane containing the other two.

$$\bar{a}, \bar{b}, \bar{c}$$ are mutually perpendicular unit vectors and $$\bar{d}$$ is a unit vector equally inclined to each other of $$\bar{a}, \bar{b}$$ and $$\bar{c}$$ at an angle of $$60^o$$. Then $$|\bar{a}+\bar{b}+\bar{c}+\bar{d}|^2=?$$

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$$4$$
0%
$$5$$
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$$6$$
0%
$$7$$

Explanation

$${\textbf{Step -1: Solve the given equations}}{\textbf{.}}$$

$$\left| {\overline a } \right| = \left| {\overline b } \right| = \left| {\overline c } \right| = \left| {\overline d } \right| = 1$$

$${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c + \overrightarrow d } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + {\left| {\overrightarrow d } \right|^2} + 2\left( {\overrightarrow {a.} \overrightarrow b + \overrightarrow {a.} \overrightarrow {c.} + \overrightarrow {a.} \overrightarrow {d.} + \overrightarrow {b.} \overrightarrow {c.} + \overrightarrow {b.} \overrightarrow {a.} + \overrightarrow {c.} \overrightarrow {d.} } \right)$$

$$ = 1 + 1 + 1 + 1 + 2\left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos 90^\circ } \right] + \left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos 90^\circ } \right] + \left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow d } \right|\cos 60^\circ } \right]$$

$$ + \left[ {\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\cos 90^\circ } \right] + \left[ {\left| {\overrightarrow b } \right|\left| {\overrightarrow d } \right|\cos 60^\circ } \right] + \left[ {\left| {\overrightarrow a } \right|\left| {\overrightarrow d } \right|\cos 60^\circ } \right]$$

$$ = 4 + 2\left[ {0 + 0 + 1 \times \dfrac{1}{2} + 0 + 1 \times 1 \times \dfrac{1}{2} + 1 \times 1 \times \dfrac{1}{2}} \right]$$

$$ = 4 + 2\left[ {\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}} \right]$$

$$ = 4 + 2 \times \dfrac{3}{2}$$

$$ = 4 + 3$$

$$ = 7$$

$${\textbf{Hence, the value of }}{\mathbf{\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c + \overrightarrow d } \right|^2} = 7.}$$

Given unit vectors $$\hat {m}, \hat {n}$$ and $$\hat {p}$$ such that $$\left( \widehat { \hat { m } \hat { n } } \right) =\hat { p } \widehat { } \left( \hat { m } \times \hat { n } \right) =\alpha$$ then the value of $$[\hat { n } \hat { p } \hat { m } ]$$ in terms of $$\alpha$$ is :

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$$\sin \alpha$$
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$$\cos \alpha$$
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$$\sin \alpha \cos \alpha $$
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$$\sin^{2} \alpha$$

$$\left| {\overline x } \right| = \left| {\overline y } \right| = 1,\,\overline x \bot \overline y ,\,\left| {\overline x + \overline y } \right| = $$

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$$\sqrt 3 $$
0%
$$\sqrt 2 $$
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$$1$$
0%
$$0$$

Which of the following is not essential for the three vectors to zero resultant?

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The resultant of any two vectors should be equal and opposite to the third vector
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They should lie in the same plane
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They should act the sides of a parallelogram
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It should be possible to represent them by the three sides of triangle taken in same order

The position vector of a point $$C$$ with respect to $$B$$ is $$\hat { i } + \hat { j }$$ and that of B with respect to A is $$\hat { i } - \hat { j }$$. The position vector of $$C$$ with respect to $$A$$ is

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$$\hat { 2i }$$
0%
$$-\hat { 2i }$$
0%
$$\hat { 2j }$$
0%
$$-\hat { 2j }$$

If $$\vec{a}, \vec{b}, \vec{c}$$ are unit vectors, then the value of $$|\vec{a}-2\vec{b}|^2+|\vec{b}-2\vec{c}|^2+|\vec{c}-2\vec{a}|^2$$ does not exceed to?

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$$9$$
0%
$$12$$
0%
$$18$$
0%
$$21$$

D,E and F are the mid-points of the sides BC,CA and AB respectively of $$\Delta ABC$$ and G is the centroid of the triangle, then $$\vec{GD}+\vec{GE}+\vec{GF}=$$

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$$\vec{0}$$
0%
$$2\vec{AB}$$
0%
$$2\vec{GA}$$
0%
$$2\vec{GC}$$

The vector $$\bar { i } +x\bar { j } +3\bar { k } $$ is rotated through an angle $$\theta $$ and doubled in magnitude, then it becomes $$4\bar { i } +\left( 4x-2 \right) \bar { j } +2\bar { k } $$. The value of x is _________.

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$$\left\{ -\frac { 2 }{ 3 } ,2 \right\} $$
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$$\left\{ \frac { 1 }{ 3 } ,2 \right\} $$
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$$\left\{ \frac { 2 }{ 3 } ,2 \right\} $$
0%
$$\left\{ 2,7 \right\} $$

Forces $$3 \vec { OA } $$, $$5 \vec { OB } $$ act along OA and OB. If their resultant passes through C on AB, then :

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C is mid-point of AB
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C divides AB in the ratio $$2:1$$
0%
$$3AC=5CB$$
0%
$$2AC=3CB$$

If $$\vec a = 4 i + 5 j - k , \vec { b } = i - 4 j + 5 k , \vec c = 3 i + j - k$$ such that $$\vec { p } \perp \vec { a } , \vec { p } \perp \overline { b }$$ and $$\vec { p . c } = 21 ,$$ then $$p =$$

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$6$( i + j + k )$$
0%
$7$( i + j + k )$$
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$6$( i - j - k )$$
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$7$( i - j - k )$$

The position vector of two points A and B are 6a+2b and a-3b. If a point C divides AB in the ratio 3 : 2, then the position vector of C is

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3a-b
0%
3a+b
0%
a+b
0%
a-b

$$\vec { a } =2\hat { I } +\hat { k } ,\vec { b } ={ b }_{ 1 }\hat { I } +{ b }_{ 2 }\hat { J+ } { b }_{ 3 }\hat { k, } \vec { a } \times \vec { b } =5\hat { I } +2\hat { J } -12\hat { k, } \hat { a } \hat { b } =11\quad then\quad { b }_{ 1 }+{ b }_{ 2 }+{ b }_{ 3 }=$$

0%
3
0%
5
0%
7
0%
9

The X & Y Component of vector A have numerical values 6 each & that of ( A +B) have numerical values 10 and 9 .What is the numerical value of B ?

0%
2
0%
3
0%
4
0%
5

If $$a,b$$ and $$c$$ are position vector of $$A,B$$ and $$C$$ respectively of $$\triangle ABC$$ and if $$|a-b|=4,|b-c|=2, |c-a|=3$$, then the distance between the centroid and incentre of $$\triangle ABC$$ is

0%
$$1$$
0%
$$\dfrac{1}{2}$$
0%
$$\dfrac{1}{3}$$
0%
$$\dfrac{2}{3}$$

Vector equation of the plane $$\vec{r}=\hat{i}-\hat{j}+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}+2\hat{j}+3\hat{k})$$ in the scalar dot product from is

0%
$$\vec{r}.(5\hat{i}-2\hat{j}+3\hat{k})=7$$
0%
$$\vec{r}.(5\hat{i}-2\hat{j}-3\hat{k})=7$$
0%
$$\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=7$$
0%
$$\vec{r}.(5\hat{i}+2\hat{j}+3\hat{k})=7$$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 12 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 12 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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