MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 8

Find $$k$$ if magnitude of vectors joining $$(0,k,0)$$ and $$(1,1,1)$$ is $$\sqrt {11}$$

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$$-3$$
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$$2$$
0%
$$-2$$
0%
$$0$$

Let $$ABC$$ be an acute scalene triangle, and $$O$$ and $$H$$ be its circumcentre and orthocentre respectively. Further let $$N$$ be the midpoint of $$OH$$. The value of the vector sum $$\vec {NA} + \vec {NB} + \vec {NC}$$ is

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$$\vec {0}$$ (zero vector)
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$$\vec {HO}$$
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$$\dfrac {1}{2}\vec {HO}$$
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$$\dfrac {1}{2}\vec {OH}$$

The projection of the vector $$\hat {i} - 2\hat {j} + \hat {k}$$ on the vector $$4\hat {i} - 4\hat {j} + 7\hat {k}$$ is

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$$\dfrac {5}{19}\sqrt {5}$$
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$$\dfrac {19}{9}$$
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$$\dfrac {9}{19}$$
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$$\dfrac {1}{19}\sqrt {6}$$

The position vectors of A, B are a, 6 respectively. The position vector of C is $$\dfrac {5\bar{a}}{3} -\bar{b}$$. Then 3

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C is inside the $$\Delta OAB $$
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C is outside the $$\Delta OAB $$ but inside the angle OAB
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C is outside the$$\Delta OAB $$ but inside the angle OBA
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None of these

$$\overline { a } ,\overline { b } ,\overline { c } $$ are three vectors such that $$\left| \overline { a } \right| =1,\left| \overline { b } \right| =2,\left| \overline { c } \right| =3$$ and $$\overline { b } ,\overline { c } $$ are perpendicular. IF projection of $$\overline { b } $$ on $$\overline { a } $$ is the same as the projection of $$\overline { c } $$ on $$\overline { a } $$, then $$\left| \overline { a } -\overline { b } +\overline { c } \right| $$

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$$\sqrt { 2 } $$
0%
$$\sqrt { 7 } $$
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$$\sqrt { 14 } $$
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$$\sqrt { 21 } $$

If $$\bar{a},\bar{b}, \bar{c}$$ are unit vectors such that $$\bar{a}+ \bar{b}+ \bar{c}=\bar{0}$$, then $$\bar{a}.\bar{b}+ \bar{b}.\bar{c}+ \bar{c}.\bar{a}=$$

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

Let $$\vec {a} = x\hat {i} + 12\hat {j} - \hat {k}, \vec {b} = 2\hat {i} + 2x\hat {j} + \hat {k}$$ and $$\vec {c} = \hat {i} + \hat {k}$$. If ordered set $$[\vec {b} \vec {c} \vec {a}]$$ is left handed, then.

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$$x\epsilon (2, \infty)$$
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$$x\epsilon (-\infty, -3)$$
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$$x\epsilon (-3, 2)$$
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$$x\epsilon \left \{3, 2\right \}$$

If $$\vec {a}$$ and $$\vec {b}$$ are unit vectors, then angle between $$\vec {a}$$ and $$\vec {b}$$ for $$\sqrt {3} \vec {a} - \vec {b}$$ to be unit vector is

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$$60^{\circ}$$
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$$90^{\circ}$$
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$$45^{\circ}$$
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$$30^{\circ}$$

Which is a unit vector?

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$$(Cos \alpha, 2 Sin \alpha)$$
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$$(Sin \alpha, Cos \alpha)$$
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$$(1, -1)$$
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$$(2Cos \alpha, Sin \alpha)$$

If $$ \vec a+2\vec b+3 \vec c =0$$, then $$\vec{b}\times \vec{c}+\vec{c}\times \vec{a}+\vec{a} \times \vec{b}$$ equals to:

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$$6\left( b\times c \right)$$
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$$\left( a\times b \right)$$
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$$6\left( c\times a \right)$$
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0

Let $$\overline a ,\overline b ,\overline c $$ be vectors of length $$3, 4, 5 $$respectively. Let $$\overline a $$ be perpendicular to $$\overline b + \overline c ,\overline b \,to\,\overline c + \overline a \,{\text{and}}\,\overline c \,{\text{to}}\,\overline a + \overline b .\,{\text{Then}}|\overline a + \overline b + \overline c |$$ is equals to:

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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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$$5\sqrt 2 $$

If the projection of $$\vec{a}$$ on $$\vec{b}$$ and the projection of $$\vec{b}$$ on $$\vec{a}$$ are equal then the angle between $$\vec{a}+\vec{b}$$ and $$\vec{a}-\vec{b}$$ is

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$$\large{\frac{\pi}{3}}$$
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$$\large{\frac{\pi}{2}}$$
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$$\large{\frac{\pi}{4}}$$
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$$\large{\frac{2\pi}{3}}$$

The value of $$|\overrightarrow A + \overrightarrow B - \overrightarrow C + \overrightarrow D |$$ can be zero if :-

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$$|\overrightarrow A | = 5,\,|\overrightarrow B | = 3,|\overrightarrow C | = 4;|\overrightarrow D | = 12$$
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$$|\overrightarrow A | = 2\sqrt 2 ,\,|\overrightarrow B | = 2,|\overrightarrow C | = 2;|\overrightarrow D | = 5$$
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$$|\overrightarrow A | = 2\sqrt 2 ,\,|\overrightarrow B | = 2,|\overrightarrow C | = 2;|\overrightarrow D | = 10$$
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$$|\overrightarrow A | = 5,\,|\overrightarrow B | = 4,|\overrightarrow C | = 3;|\overrightarrow D | = 8$$

Let $$\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}, \,\,\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$$ and $$\vec{c} = \vec{i} + \vec{j} - 2\vec{k} $$ be three vectors. A vector of the type $$\vec{b} + \lambda \vec{c}$$ for some scalar $$\lambda$$, whose projection on $$\vec{a}$$ is of magnitude $$\sqrt {\frac{2}{3}}$$. Thenthe value of $$\lambda$$ is

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$$1$$
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$$0$$
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$$-1$$
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$$2$$

Given that P = 12, Q = 5 and R = 13 also $$\vec P + \vec Q = \vec R$$, then the angle between $$\vec P$$ and $$\vec Q$$ will be :

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$$

{\pi}

$$
0%
$$

{\pi}\over{2}

$$
0%
zero
0%
$$

{\pi}\over{4}

$$

The vertices of a triangle are $$A(1,1,2),B(4,3,1)$$ and $$C(2,3,5).$$ A vector representing the internal bisector of the angle $$A$$ id

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

A unit vector along the direction $$\hat i + \hat j + \hat k$$ has a magnitude:

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

If $$\overline {OA}=i+j+k, \overline {AB}=3i-2j+k,\overline {BC}=i+2j-2k$$ and $$\overline {CD}=2i+j+3k $$ then find the vector $$\overline{OD}$$.

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$$7i-2j-6k$$
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$$7i+2j+3k$$
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$$7i+2j+5k$$
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None of these

Let $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ and $$\vec{d}$$ are four distinct vectors satisfying the conditions $$\vec{a} \times \vec{b}$$ = $$\vec{c} \times \vec{d}$$ & $$\vec{a} \times \vec{c}$$ =$$\vec{b} \times \vec{d}$$ then $$\vec{a}.\vec{b} + \vec{c}.\vec{d} \neq \vec{a}.\vec{c} + \vec{b}.\vec{d}$$ .

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True
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False

Four forces act on a point object. The object will be in equilibrium, if:

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all of them are in the same plane
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they are opposite to each other in pairs
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the sum of x, y and z - components of forces zero separately
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they form a closed figure of 4 sides when added as Polygon law

The value of $$\bar {a} \times (\bar {b}+\bar {c})+\bar {b}\times (\bar {c}+\bar {a})+\bar {c}\times (\bar {a}+\bar {b})=$$

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$$0$$
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$$-1$$
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$$2\ \bar {a}\times (\bar {b}+\bar {c})$$
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$$[\bar {a}\bar {b}\bar {c}]$$

Which of the following is the unit vector perpendicular to $$ \vec{A} $$ and $$ \vec{B} $$ ?

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$$ {{\vec{A}\times \vec{B}} \over {AB\sin \theta }}$$
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$$\left|{{\vec{A}\times \vec{B}} \over {AB\sin \theta }}\right|$$
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$$ {{\vec{A}\times \vec{B}} \over {AB\cos \theta }}$$
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$$\left|{{\vec{A}\times \vec{B}} \over {AB\cos \theta }}\right|$$

The vector that must be added to the vector $$\hat{i}-3\hat{j}+2\hat{k}$$ and $$3\hat{i}+6\hat{j}-7\hat{k}$$ so that the resultant vector is a unit vector along the y-axis is:

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$$4\hat{i}+2\hat{j}+5\hat{k}$$
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$$-4\hat{i}-2\hat{j}+5\hat{k}$$
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$$3\hat{i}+4\hat{j}+5\hat{k}$$
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Null vector

The vector equation of the plane containing the line $$\vec {r}=(-2\hat {i}-3\hat {j}+4\hat {k})+\lambda(3\hat {i}-2\hat {j}-\hat {k})$$ and the point $$\hat {i}+2\hat {j}+3\hat {k}$$ is:

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$$\vec {r}\cdot(\hat {i}+3\hat {k})=10$$
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$$\vec {r}\cdot(\hat {i}-3\hat {k})=10$$
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$$\vec {r}\cdot(3\hat {i}+\hat {k})=10$$
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$$none\ of\ these$$

Let $$\vec{a},\vec{b}$$ and $$\vec{c}$$ be three non-zero vectors such that no two of these are collinear. If the vector $$\vec{a}+2\vec{b}$$ is collinear with $$\vec{c}$$ and $$\vec{b}+3\vec{c}$$ is collinear with $$\vec{a}(\lambda$$ being some non-zero scalar), then $$\vec{a}+2\vec{ b}+6\vec{c}$$ equals

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$$\lambda\vec{a}$$
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$$\lambda\vec{b}$$
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$$\lambda\vec{c}$$
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$$\vec{0}$$

Component of $$\vec{a}=\hat{i}-\hat{j}-\hat{k}$$ perpendicular to the vector $$\vec{b}=2\hat{i}+\hat{j}-\hat{k}$$ is?

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$$\dfrac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$$
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$$\dfrac{1}{3}(\hat{i}-4\hat{j}-2\hat{k})$$
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$$\dfrac{1}{3}(\hat{i}+4\hat{j}+2\hat{k})$$
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$$\dfrac{1}{3}(\hat{i}+2\hat{j}+\hat{k})$$

If $$\vec{a}, \vec{b}, \vec{c}$$ are three non-coplanar vectors such that $$\vec{d}\cdot \vec{a}=\vec{d}\cdot \vec{b}=\vec{d}\cdot \vec{c}=0$$, then $$\vec{d}$$ is :-

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a scalar
0%
a null vector
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has non-zero magnitude
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an imaginary number

$$\overline a = \overline i - \overline k ,\overline b = x\overline i + \overline j + (1 - x)\overline k $$ and $$\overline c = y\overline i + \lambda \overline j + (1 + x - y)\overline k $$ then $$\left[ {\overline a \overline b \overline c } \right]$$ depends on:-

0%
neither x nor y
0%
both x and y
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only x
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only y

$$A(1, -1, -3)$$, $$B(2, 1, -2)$$ & $$C(-5, 2, -6)$$ are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is?

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$$\sqrt{10}/4$$
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$$3\sqrt{10}/4$$
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$$\sqrt{10}$$
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None

If $$\hat {i},\hat {j},\hat {k}$$ are positive vectors of $$A,B,C$$ and $$\vec {AB}=\vec {CX}$$, then positive vector of $$X$$ is

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$$-\hat {i}+\hat {j}+\hat {K}$$
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$$\hat {i}-\hat {j}+\hat {K}$$
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$$\hat {i}+\hat {j}-\hat {K}$$
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$$\hat {i}+\hat {j}+\hat {K}$$

The $$P.V.'s$$ of the vertices of a $$\triangle ABC$$ are $$\bar {i}+\bar {j}+\bar {k}, 4\bar {i}+\bar {j}+\bar {k}, 4\bar {i}+5\bar {j}+\bar {k}$$. The $$P.V.$$ of the circumcentre of $$\triangle ABC$$ is

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$$\dfrac{5}{2}\bar {i}+3\bar {j}+\bar {k}$$
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$$5\bar {i}+\dfrac{3}{2}\bar {j}+\bar {k}$$
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$$5\bar {i}+3\bar {j}+\dfrac{1}{2}\bar {k}$$
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$$\bar {i}+\bar {j}+\bar {k}$$

The position vector of point P, is?

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$$\dfrac{3|\vec{a}||\vec{c}|}{3|\vec{c}|+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$$
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$$\dfrac{|\vec{a}||\vec{c}|}{3|\vec{c}+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$$
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$$\dfrac{2|\vec{a}||\vec{c}|}{3|\vec{c}|+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\bar{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$$
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$$\dfrac{3|\vec{a}||\vec{c}|}{3|\vec{c}|+2\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}-\dfrac{\vec{c}}{|\vec{c}|}\right\}$$

Explanation

According to question:

$$\begin{array}{l} B=\overrightarrow { a } +\overrightarrow { c } \\ \overrightarrow { E } =\dfrac { { 2\overrightarrow { c } +\overrightarrow { a } +\overrightarrow { c } } }{ 3 } =\dfrac { { 3\overrightarrow { c } +\overrightarrow { a } } }{ 3 } \\ then, \\ Position\, vector\, is: \\ \, \, \, \, \, \overrightarrow { OP } =\lambda \, (\overrightarrow { a } +\overrightarrow { c } )-----(i) \\ \, \, \, \, Now, \\ \, \, \, \, \, \, \, positon\, \, vector\, of\overrightarrow { \, p } : \\ \, \, \, \, \, \overrightarrow { p } =\overrightarrow { a } +\mu \left( { \dfrac { { 3\overrightarrow { c } +\overrightarrow { a } } }{ 3 } -\overrightarrow { a } } \right) \\ \, \, \, \overrightarrow { p } =\overrightarrow { a } +\mu \left( { \dfrac { { 3\overrightarrow { c } -2\overrightarrow { a } } }{ 3 } } \right) -----(ii) \\ Now,\, equating\, \, eqn.\, \, (i)\, \, \& \, (ii) \\ \lambda \, \left( { \dfrac { { \overrightarrow { a } } }{ { \left| { \overline { a } } \right| } } +\dfrac { { \overrightarrow { c } } }{ { \left| { \overline { c } } \right| } } } \right) =\overrightarrow { a } +\dfrac { \mu }{ 3 } \left( { 3\overrightarrow { c } -2\overrightarrow { a } } \right) \\ \dfrac { \lambda }{ { \left| { \overline { a } } \right| } } =1-\dfrac { { 2\mu } }{ 3 } \, \, \, \, \, \, \, (cofficient\, compare) \\ \dfrac { \lambda }{ { \left| { \overline { a } } \right| } } =\dfrac { { 3\mu } }{ 3 } ,\, \, \, \, \, \, \, \, \, \mu =\dfrac { \lambda }{ { \left| { \overline { a } } \right| } } \\ And, \\ \dfrac { \lambda }{ { \left| { \overline { a } } \right| } } =1-\dfrac { 2 }{ 3 } .\, \dfrac { \lambda }{ { \left| { \overline { c } } \right| } } \\ \lambda \left( { \dfrac { 1 }{ { \left| { \overline { a } } \right| } } +\dfrac { 2 }{ 3 } .\, \dfrac { 1 }{ { \left| { \overline { c } } \right| } } } \right) =1 \\ \lambda =\dfrac { { 3\left| { \overline { a } } \right| \, \left| { \overline { c } } \right| } }{ { 3\left| { \overline { c } } \right| +2\left| { \overline { a } } \right| } } \\ \, positon\, \, vector\, of\overrightarrow { \, p } : \\ \overrightarrow { p } =\dfrac { { 3\left| { \overline { a } } \right| \, \left| { \overline { c } } \right| } }{ { 3\left| { \overline { c } } \right| +2\left| { \overline { a } } \right| } } \, \, \, \left( { \dfrac { { \overrightarrow { a } } }{ { \left| { \overline { a } } \right| } } +\dfrac { { \overrightarrow { c } } }{ { \left| { \overline { c } } \right| } } } \right) \\ so\, ,\, the\, correct\, \, option\, is\, A. \end{array}$$

If $$a^b=b^c=ab$$, then $$b+c$$ always equals?

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

If $$\overline { a } =\cfrac { 1 }{ \sqrt { 10 } } \left( 3\overline { i } +\overline { k } \right) ;\overline { a } =\cfrac { 1 }{ 7 } \left( 2\overline { i } +3\overline { j } -6\overline { k } \right) $$ then the value of
$$(2\overline { a } -\overline { b } ).[(\overline { a } \times \overline { b } )\times (\overline { a } +2\overline { b } )]$$

0%
$$5$$
0%
$$3$$
0%
$$-5$$
0%
$$-3$$

Let P, Q, R and S be the points on the plane with position vectors $$-2\hat{i}-\hat{j}, 4\hat{i}, 3\hat{i}+3\hat{j}$$ and $$-3\hat{i}+2\hat{j}$$ respectively. the quadrilateral PQRS must be a.

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Parallelogram, which is neither a rhombus nor a rectangle
0%
Square
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Rectangle, but not a square
0%
Rhombus, but not a square

If $$\vec a,\vec b,\vec c$$ non-zero vectors such that $$\vec a$$ is perpendicular to $$\vec b$$ and $$\vec c$$ and non-zero vector coplanar with $$\vec a + \vec b$$ and $$2\vec b - \vec c$$ and $$\vec d.\vec a = 1$$ , then the minimum value of $$\left| {\vec d} \right|$$

0%
$$\frac{2}{{\sqrt {13} }}$$
0%
$$\frac{1}{{\sqrt {13} }}$$
0%
$$\frac{3}{{\sqrt {13} }}$$
0%
$$\frac{4}{{\sqrt {13} }}$$

If $$a,b,c \in N$$, the number of points having position vectors $$a\hat i + b\hat j + c\hat k$$ such that $$6 \le a + b + c \le 10$$ is

0%
110
0%
116
0%
120
0%
127

If $$\bar a + \bar b$$ is perpendicular to $$\bar b$$ and $$\bar a + 2\bar b$$ is perpendicular to $$\bar a$$ then.

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$$\left| {\bar a} \right| = \left| {\bar b} \right|$$
0%
$$\left| {\bar a} \right| = \sqrt 2 \left| {\bar b} \right|$$
0%
$$\left| {\bar b} \right| = \sqrt 2 \left| {\bar a} \right|$$
0%
$$\left| {\bar a} \right| = \left| {\bar b} \right|\sqrt 3 $$

The length of the projection of the line segment joining points $$(5, -1, 4)$$ and $$(4, -1, 8)$$ on the plane $$x+y+z=7$$.

0%
$$\dfrac{2}{\sqrt{3}}$$
0%
$$\dfrac{2}{3}$$
0%
$$\sqrt{5}$$
0%
$$\sqrt{\dfrac{2}{3}}$$

A unit vector $${\vec a}$$ in the plane of $$\vec = 2\hat i + \hat j$$ and $$\vec c = \hat i - \hat j + \hat k$$ is such that angle between $${\vec a}$$ and $${\vec b}$$ is the same angle between $${\vec a}$$ and $${\vec d}$$ where $$\vec d = \hat j + 2\hat k$$

0%
$$\frac{{\hat i + \hat j + \hat k}}{{\sqrt 3 }}$$
0%
$$\frac{{\hat i - \hat j + \hat k}}{{\sqrt 3 }}$$
0%
$$\frac{{2\hat i + \hat j}}{{\sqrt 5 }}$$
0%
$$\frac{{2\hat i - \hat j}}{{\sqrt 5 }}$$

If the unit vectors $$\vec{e}_1 \, and \, \vec{e}_2$$ are inclined at an angle $$2 \theta \, and \, |\vec{e}_1 - \vec{e}_2| < 1$$, then for $$\theta \in [0, \pi] , \theta$$ may lie in the interval

0%
$$\left[0, \dfrac{\pi}{6} \right]$$
0%
$$\left[ \dfrac{\pi}{6} , \dfrac{\pi}{2} \right]$$
0%
$$\left[ \dfrac{5 \pi}{6} , \pi \right]$$
0%
$$\left[\dfrac{\pi}{2}, \dfrac{5 \pi}{6} \right]$$

Let $$\vec{a},\vec{b},\vec{c}$$ be three non-zero vectors such that $$\vec{a}+\vec{b}+\vec{c}=0$$ and $$\lambda \vec{b}\times \vec{a}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}=0$$, then $$\lambda$$ is

0%
$$1$$
0%
$$2$$
0%
$$-1$$
0%
$$-2$$

If $$[\vec{a}\vec{b}\vec{c}]=2$$, then find the value of $$[(\vec{a}+2\vec{b}-\vec{c}(\vec{a}-\vec{b})(\vec{a}-\vec{b}-\vec{c})]$$.

0%
$$6$$
0%
$$8$$
0%
$$12$$
0%
$$11$$

A non-zero vectors $$\overrightarrow{a}$$ is such that its projections along the vectors $$\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}$$ and $$\dfrac{-\hat{i}+\hat{j}}{\sqrt{2}}$$ and $$\hat{k}$$ are equal then unit vector along $$\overrightarrow{a}$$ is

0%
$$\dfrac{\sqrt{2}\hat{j}-\hat{k}}{\sqrt{3}}$$
0%
$$\dfrac{\hat{j}-\sqrt{2}\hat{k}}{\sqrt{3}}$$
0%
$$\dfrac{\sqrt{2}}{\sqrt{3}}\hat{j}+\dfrac{\hat{k}}{\sqrt{3}}$$
0%
$$\dfrac{\hat{j}-\hat{k}}{\sqrt{2}}$$

If $$a$$ and $$b$$ are unit vectors along $$OA, OB$$ and $$OC$$ bisects the angle $$AOB$$. The unit vector along $$OC$$ is

0%
$$\dfrac{a+b}{2}$$
0%
$$\dfrac{b-b}{2}$$
0%
$$\dfrac{a+b}{|a+b|}$$
0%
$$\dfrac{a-b}{|a-b|}$$

If $$\overrightarrow {a}$$, $$\overrightarrow {b}$$ and $$\overrightarrow {c}$$ be three non-zero vectors, non-coplanar and if $$\overrightarrow {d}$$ is such that $$\bar { a } =\dfrac { 1 }{ y } \left( \overrightarrow { b } +\overrightarrow { c } +\overrightarrow { d} \right) $$ and where $$x$$ and $$y$$ are non-zero real numbers, then $$\dfrac { 1 }{ xy } \left( \vec { a } +\vec { b } +\vec { c } +\vec { d } \right) =$$

0%
$$-\vec {a}$$
0%
$$\vec {0}$$
0%
$$-2\vec {a}$$
0%
$$-3\vec {c}$$

If the vector $$6\hat { i } -3\hat { j } -6\hat { k } $$ is decomposed into vectors parallel and perpendicular to the vector $$\hat { i } +\hat { j } +\hat { k }$$ then the vectors are :

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$$-\left( \hat { i } +\hat { j } +\hat { k } \right) +\hat { 7i } -\hat { 2j } -\hat { 5k } $$
0%
$$-2\left( \hat { i } +\hat { j } +\hat { k } \right) +\hat { 8i } -\hat { j } -\hat { 4k } $$
0%
$$\left( \hat { i } +\hat { j } +\hat { k } \right)+ \hat { 4i } -\hat { 5j } -\hat { 8k } $$
0%
$$none$$

The value of $$\left(a.i\right)i+\left(a.j\right)j+\left(a.k\right)k$$ in terms of vector $$a$$

0%
$$\overrightarrow{a}$$
0%
$$\overrightarrow{a}\hat{i}$$
0%
$$\overrightarrow{a}\hat{j}$$
0%
$$\overrightarrow{a}\hat{k}$$

If the position vectors of A, B, C, D are $$\vec{a}, \vec{b}. 2\vec{a}+3\vec{b}, \vec{a}-2\vec{b}$$ respectively, then $$\vec{AC}, \vec{DB}, \vec{BA}, \vec{DA}$$ are?

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$$\vec{a}+3\vec{b}, 3\vec{b}-\vec{a}, \vec{a}-\vec{b}, 2\vec{b}$$
0%
$$2\vec{b}, \vec{b}-2\vec{a}, 3\vec{b}+\vec{a}, \vec{b}-\vec{a}$$
0%
$$\vec{a}-3\vec{b}, 3\vec{b}-\vec{a}, \vec{a}+\vec{b}, 2\vec{b}$$
0%
$$-2\vec{b}, \vec{b}-2\vec{a}, 3\vec{b}-\vec{a}, \vec{b}-\vec{a}$$

What vector must be added to the two vectors $$\hat{i}+2\hat{j}+2\hat{k}$$ and $$2\hat{i}-\hat{j}-\hat{k}$$, so that the resultant may be a unit vector along x-axis

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

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 8 - 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 8 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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