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CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 13 - MCQExams.com
CBSE
Class 12 Commerce Maths
Vector Algebra
Quiz 13
If $$\vec { u } =\vec { a } -\vec { b } ~;\vec{ v } =\vec { a } +\vec { b } ~~\&~ |\vec { a }| =|\vec { b }| =2,$$ then $$\left| \vec { u } \times \vec { \upsilon } \right| $$ is equal to:
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$$\sqrt { 2\left( 4-\left( \vec { a } .\vec { b } \right) ^{ 2 } \right) } $$
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$${ 2 }\sqrt { \left( 16+\left( \vec { a } .\vec { b } \right) ^{ 2 } \right) } $$
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$${ 2 }\sqrt { \left( 4-\left( \vec { a } .\vec { b } \right) ^{ 2 } \right) } $$
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$${ 2 }\sqrt { \left( 16-\left( \vec { a } .\vec { b } \right) ^{ 2 } \right) } $$
$$\vec{r}.\hat{i}=2\vec{r}.\hat{j}=4\vec{r}.\hat{k}$$ and $$\left|\vec{r}\right|=\sqrt{84}$$, then $$\left|\vec{r}.\left(2\hat{i}-3\hat{j}+\hat{k}\right)\right|$$ is equal to
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$$0$$
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$$2$$
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$$4$$
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$$6$$
If $$\bar{a} = \hat{i} + \hat{j} - 2 \hat{k}, \bar{b} = 2 \hat{i} - 0 \hat{j} + \hat{k}, \bar{c} = 3 \hat{i} - \hat{k}$$ and $$\bar{c} = m \bar{a} + n \bar{b}$$ then m + n = ....
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0
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1
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2
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-1
If the position vectors of $$A, B, C, D$$ are $$3\hat{i} + 2\hat{j} + \hat{k}, 4\hat{i} + 5\hat{j} + 5\hat{k}, 4\hat{i} + 2\hat{j} - 2\hat{k}, 6\hat{i} + 5\hat{j} - \hat{k}$$ respectively then the position vector of the point of intersection of $$\bar{AB}$$ and $$\bar{CD}$$ is
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$$2\hat{i} + \hat{j} - 3\hat{k}$$
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$$2\hat{i} - \hat{j} + 3\hat{k}$$
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$$2\hat{i} + \hat{j} + 3\hat{k}$$
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$$2\hat{i} - \hat{j} - 3\hat{k}$$
Explanation
$$ \begin{aligned} \overrightarrow{A B} &=\vec{b}-\vec{a} \\ &=(4 i+5 \hat{\jmath}+5 \hat{k})-(3 \hat{i}+2 \hat{\jmath}+\hat{k}) \\ &=\hat{\imath}+3 \hat{\jmath}+4 \hat{k} , and \end{aligned} $$
$$\begin{aligned} \overrightarrow{C D} &=\vec{d}-\vec{c} \\ &=(6 \hat{\imath}+5 \hat{\jmath}-\hat{k})-(4 \hat{\imath}+2 \hat{\jmath}-2 \hat{k}) \\ &=\overrightarrow{2} \hat{\imath}+3 \hat{\jmath}+\hat{k} \\ \text { Let } \overrightarrow{A B} & \text { and } \overrightarrow{C D} \text { intersect at } \vec{P} . \\ \text { Line } A B=(3 i+2 j+\hat{k})+\lambda(i+3 \hat{j}+4 \hat{k}) & \text { and } \end{aligned}$$
Line $$C D=(4 i+2 \hat{\imath}-2 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$$
W.r.t. line $$A_{0}$$,
$$P=(3+\lambda, 2+3 \lambda, 1+4 \lambda)$$
and $$w \cdot r \cdot t \cdot$$ line $$C D$$,
$$P=(4+2 \mu, 2+3 \mu,-2+\mu)$$
Comparing both points,
$$\begin{aligned} \Rightarrow & 3+\lambda^{\prime}=4+2 \mu, \\ & 2+3 \lambda=2+3 \mu \\ & 1+4 \lambda=-2+\mu \end{aligned}$$
Solving these equs.,
$$\therefore \vec{p}=2 \hat{\imath}-\hat{\jmath}-3 \hat{k}$$
$$\therefore$$ option $$D$$ is correct.
Let $$\vec{a}=\hat{i}-\hat{j},\vec{b}=\hat{i}-\hat{j}=\vec{c}=\hat{i}-\hat{j}$$, if $$\vec{d}$$ is a unit vector such that $$\vec{a}.\vec{d}=0=|\vec{b}\vec{c}\vec{d}|$$ then $$\vec{d}$$ equals:
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$$\pm\dfrac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}$$
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$$\pm\dfrac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{3}}$$
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$$\pm\dfrac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{3}}$$
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$$\pm\hat{k}$$
If $$a=i-j,b=i+j,c=i+3j+5k$$ and $$n$$ is a unit vector such that $$b,n=0,a,n=0$$ then the value of $$|c,n|$$ is equal to
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$$1$$
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$$3$$
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$$5$$
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$$2$$
Explanation
$$\begin{array}{l} { b_{ n } }=0 \\ \left( { \hat { i } +\hat { j } } \right) \left( { a\hat { i } +\hat { i } e } \right) =0 \\ a+b=c \\ \left( { -\hat { j } } \right) \left( { a\hat { i } +b\hat { j } } \right) =0 \\ a-b=0\, \, \, \, \, \, \, a=0=b \\ \left( { \bar { c } \cdot \bar { n } } \right) \\ \left( { \hat { i } +3\hat { j } +5\hat { b } } \right) \cdot \left( { c\hat { k } } \right) \\ =5c\, =\sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } =1 \\ =5 \end{array}$$
The value of $$|\vec{a} \times \hat{i} |^2 + |\vec{a} \times \hat{j} |^2 + |\vec{a} \times \hat{k} |^2$$ is
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$$a^2$$
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$$2a^2$$
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$$3a^2$$
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none of these
The values of $$\lambda$$ such that $$(x, y, z) \neq (0, 0, 0)$$ and $$(\hat{i} + \hat{j} + 3\hat{k})x + (3\hat{i} - 3\hat{j} + \hat{k})y + (-4\hat{i} + 5\hat{j})z = \lambda(x\hat{i} + y\hat{j} + z\hat{k})$$ are
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$$0, 1$$
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$$0, -1$$
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$$1, -1$$
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$$0, 1, -1$$
Explanation
$$\Rightarrow(x+3 y-4 z-\lambda x)\hat{i}+(x-3 y+5 z-\lambda y) \hat{j}+(3 x+y-\lambda z) \hat{k}=0$$
$$\Rightarrow x+3 y-4 z-\lambda x=0$$
$$x-3 y+5 z-\lambda y=0$$
$$3 x+y-\lambda z=0$$
$$\left[\begin{array}{ccc}1-\lambda & 3 & -4 \\ 1 & -3-\lambda & 5 \\ 3 & 1 & -\lambda\end{array}\right]\left[\begin{array}{c}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$$
$$\Rightarrow\left|\begin{array}{ccc}1-\lambda & 3 & -4 \\ 1 & -3-\lambda & 5 \\ 3 & 1 & -\lambda\end{array}\right|=0 \quad\left\{\begin{array}{c}\because \text { for non-trivial } \\ \text { soution of equation } \\ A x=y \\ |A|=0\end{array}\right\}$$
$$\Rightarrow(1-\lambda)((-3-\lambda)(-\lambda)-5)-1(-3 \lambda+4)+3(15+4(-3-\lambda))=0$$
$$=(1-\lambda)\left(3 \lambda+\lambda^{2}-5\right)-4+3 \lambda+3(3-4 \lambda)=0$$
$$=(1-\lambda)\left(\lambda^{2}+3 \lambda-5\right)-4+3 \lambda+9-12 \lambda=0$$
$$=\lambda^{2}+3 \lambda-5-\lambda^{3}-3 \lambda^{2}+5 \lambda-4+3 \lambda+9-12 \lambda=0$$
$$=-\lambda^{3}-2 \lambda^{2}-\lambda=0$$
$$\Rightarrow \lambda^{3}+2 \lambda^{2}+\lambda^{2}=0$$
$$\lambda\left(\lambda^{2}+2 \lambda+1\right)=0$$
$$\lambda=0, \quad \lambda^{2}+2 \lambda+1=0$$
$$(\lambda+1)^{2}=0$$
$$\lambda=-1$$
$$\Rightarrow \lambda=0,-1$$
$$\therefore$$ option $$B$$ is correct.
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 $$\vec{v}$$ along $$\vec{u}$$ is equal to that of $$\vec{w}$$ along $$\vec{u}$$ and $$\vec{v}$$, $$\vec{w}$$ are perpendicular to each other , then $$\left | \vec{u} \vec{v} + \vec{w} \right |$$ equals
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$$\sqrt{14}$$
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$$\sqrt{7}$$
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2
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14
Explanation
Given,
$$|\vec{u}|=1$$ and $$|\vec{v}|=2,|\vec{w}|=3$$
and, $$\quad \vec{v} \cdot \vec{u}=\vec{w} \cdot \vec{u}$$
$$\vec{v} \cdot \vec{u}=\vec{\omega} \cdot \vec{u}$$
or, $$\quad \vec{v} \cdot \vec{\omega}=0$$
Now, $$\quad|\vec{u}-\vec{v}+\vec{w}|$$
Squaring,$$|\vec{u}|^{2}+\left.\vec{v}\right|^{2}+|\vec{w}|^{2}-2 \vec{u} \cdot \vec{v}-2 \vec{v} \cdot \vec{\omega}$$
$$+2 \vec{u} \cdot \vec{w}$$
$$\Rightarrow 1+4+9-2\vec{u} \cdot\vec{v}+2 \overrightarrow{u} \cdot \vec{v}$$-0
$$\Rightarrow \quad 14$$
$$|\vec{u}-\vec{v}+\vec{w}|=\sqrt{14}$$
option $$A=\sqrt{14}$$
Vector $$\vec { x }$$ satisfying the relation $$\vec { A } . \overline { x } = c$$ and $$\vec { A } \times \vec { x } = \vec { B }$$ is
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$$\frac { c \vec { A } - ( \overline { A } \times \vec { B } ) } { | \vec { A } | }$$
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$$\frac { c \vec { A } - ( \vec { A } \times \vec { B } ) } { | \vec { A } | ^ { 2 } }$$
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$$\frac { c \vec { A } + ( \vec { A } \times \vec { B } ) } { | \vec { A } | ^ { 2 } }$$
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None
For any vector $$\vec a$$, the value of $${ (\vec a\times \hat i) }^{ 2 }+{ (\vec a\times \hat j) }^{ 2 }+{ (\vec a\times \hat k) }^{ 2 }$$ is equal to
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$$4{ |\vec a| }^{ 2 }$$
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$$2{ |\vec a| }^{ 2 }$$
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$${|\vec a |}^{ 2 }$$
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$$3{ |\vec a| }^{ 2 }$$
let $$\bar { a } $$ be a unit vector and $$\bar { b } $$ be a non-zero vector not parallel to $$\bar { a } $$ if two sides of a triangle are represented by the vectors $$\sqrt { 3 } \left( \bar { a } \times \bar { b } \right) \quad and\quad \bar { b } - \left( \bar { a } .\bar { b } \right) \bar { a } $$ then the angles of triangle are
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$$90^{ \circ },{ 60 }^{ \circ },{ 30 }^{ \circ }$$
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$$45^{ \circ },{ 45 }^{ \circ },{ 90 }^{ \circ }$$
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$$60^{ \circ },{ 60 }^{ \circ },{ 60 }^{ \circ }$$
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$$75^{ \circ },{ 45 }^{ \circ },{ 60 }^{ \circ }$$
In a triangle ABC, if $$A=(0, 0), B=(3, 3\sqrt{3}), C=(-3\sqrt{3}, 3)$$ then the vector of magnitude $$2\sqrt{2}$$ units directed along $$\overline{AO}$$, where O is the circumcentre of triangle ABC is?
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$$(1-\sqrt{3})\bar{i}+(1+\sqrt{3})\bar{j}$$
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$$\sqrt{3}\bar{i}+2\bar{j}$$
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$$\bar{i}-\sqrt{3}\bar{j}$$
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$$\bar{i}+2\bar{j}$$
In a parallelogram ABD, $$|\overset { \_ }{ A\overset { \rightarrow }{ B } } |=a,|\overset { \_ }{ A\overset { \rightarrow }{ D } } |=b$$ and $$|\overset { \_ \\ \quad \quad \rightarrow }{ AC } |=c,$$, $$\overset { \_ \rightarrow }{ AB } .\overset { \_ \rightarrow }{ DB } $$ has the value :
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$$\frac { 1 }{ 2 } ({ 3a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 })$$
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$$\frac { 1 }{ 2 } ({ a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 })$$
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$$\frac { 1 }{ 2 } ({ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 })$$
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$$\frac { 1 }{ 3 } ({ b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 })$$
If the position vectors of the vertices $$A,B$$ and $$C$$ of a $$\Delta ABC$$ are respectively $$4\hat{i}+7\hat{j}+8\hat{k},2\hat{i}+3\hat{j}+4\hat{k}$$ and $$2\hat{i}+5\hat{j}+7\hat{k}$$, then the position vector of the point, where the bisector of $$\angle A$$ meets $$BC$$ is:
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$$\frac{1}{2}(4\hat{i}+8\hat{j}+11\hat{k})$$
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$$\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$$
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$$\frac{1}{4}(8\hat{i}+14\hat{j}+19\hat{k})$$
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$$\frac{1}{3}(6\hat{i}+8\hat{j}+15\hat{k})$$
$$If\quad |\overrightarrow { a } |$$ =2 and $$\quad |\overrightarrow { b } |$$=3 and $$\quad |\overrightarrow { a } |$$.$$\quad |\overrightarrow { b } |$$ =Then (a x (a x (a x (a x b)))) is equal to
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$$4\hat { b } $$
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$$-4\hat { b }$$
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$$4\hat { a } $$
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$$-4\hat { a } $$
If C is the mid-point of AB and P is any point outside AB , then
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$$\overrightarrow{PA}$$ + $$\overrightarrow{PB}$$ + $$\overrightarrow{PC}$$ = 0
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$$\overrightarrow{PA}$$ + $$\overrightarrow{PB}$$ + 2$$\overrightarrow{PC}$$ = $$\overrightarrow{0}$$
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$$\overrightarrow{PA}$$ $$\overrightarrow{PB}$$ = $$\overrightarrow{PC}$$
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$$\overrightarrow{PA}$$ $$\overrightarrow{PB}$$ = 2$$\overrightarrow{PC}$$
If $$\overset { \rightarrow }{ a } $$ and $$\overset { \rightarrow }{ b } $$ are vectors such that $$|\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } |=\sqrt { 29 } $$ and $$\overset { \rightarrow }{ a } \times (2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +4\overset { \wedge }{ k } )=(2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +4\overset { \wedge }{ k } )\times \overset { \rightarrow }{ b } $$, then a possible value of $$(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } )(-7\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } )$$ is
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0%
0
0%
3
0%
4
0%
8
If $$\hat{u}$$ and $$\hat{v}$$ are unit vectors and $$\theta$$ is the acute angle between them , then 2$$\hat{u}$$ 3$$\hat{v}$$ is a unit vector for
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No value of $$\theta$$
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Exactly one value of $$\theta$$
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Exactly two values of $$\theta$$
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More than two values of $$\theta$$
Explanation
Given,
$$\bar{u}$$ and $$\vec{v}$$ are unit vector
$$|\vec{u}|=1$$
$$|\vec{v}|=1$$
Now,
$$\Rightarrow|2 \vec{u}||\overrightarrow{3 v}| \sin \theta=1$$
$$|\overrightarrow{2 u} \times \rightarrow{\vec{3 v}}|$$
$$[\because|\vec{a} \times \vec{b}|=|\vec a|\vec b|| \sin \theta]$$
$$\Rightarrow 2 \cdot 3|\bar{u}| \nabla \mid \sin \theta=1 \\$$
$$\sin \theta=\frac{1}{6}$$
$$\text { There is only one value of } \theta \\$$
$$\text { for which }|\vec 24 \times \vec 3V| \text { is a } \\$$
$$\text { unit vector. }$$
$$\text { option } B$$
Let $$\vec{a}$$ = $$\hat{i}$$ $$\hat{j}$$, $$\vec{b}$$ = $$\hat{j}$$ $$\hat{k}$$, $$\vec{c}$$ = $$\hat{k}$$ $$\hat{i}$$. If $$\vec{d}$$ is a unit vector such that $$\vec{a}.\vec{d}$$ = 0 = $$\left | \vec{b}\vec{c}\vec{d} \right |$$ then $$\vec{d}$$ equals :
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$$\dfrac{\hat{i} + \hat{j} 2\hat{k}}{\sqrt{6}}$$
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$$\dfrac{\hat{i} + \hat{j} \hat{k}}{\sqrt{3}}$$
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$$\dfrac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$$
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$$ \hat{k}$$
The value of $$\lambda $$ for $$\left( x,y,z \right) \neq \left( 0,0,0 \right) $$ and $$\left( i+j+3k \right) x+\left( 3i-3j+k \right) y+\left( -4i+5j \right) z=\lambda \left( xi+yj+zk \right) $$ are
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0, -1
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0, 1
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-2, 0
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0, 2
If $$\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ c } $$ are unit vectors such that $$\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } =0$$ then the value of $$\overset { \rightarrow }{ a. } \overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ a. } $$ is
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1
0%
-1
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-3/2
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none of these
If $$\overline { a } =-2\overline { i } +3\overline { j } +4\overline { k } $$ and $$\overline { b } =-2\overline { i } -2\overline { j } +3\overline { k } $$ then $$\overline { a } .\overline { b } $$ is
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2
0%
-2
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6
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none of these
Let position vector of the orthocentre of $$\triangle ABC$$ be $$\overrightarrow{r}$$. then, which of the following statement(s) is\are correct (Given position vector of points $$a\hat{i},b\hat{j},c\hat{k}$$ and $$abc=0$$)
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$$\displaystyle \overrightarrow { r } .\bar { i } =\frac{a}{\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}}}$$
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$$\displaystyle \overrightarrow { r } .\bar { i } =\frac{1}{a\left(\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}} \right) }$$
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$$\displaystyle \frac {\overrightarrow { r } .\bar { i }}{\overrightarrow { r } .\bar { j }}+\frac {\overrightarrow { r } .\bar { j }}{\overrightarrow { r } .\bar { k }}+\frac {\overrightarrow { r } .\bar { k }}{\overrightarrow { r } .\bar { i }}=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$
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$$\displaystyle \frac {\overrightarrow { r } .\bar { i }}{\overrightarrow { r } .\bar { j }}+\frac {\overrightarrow { r } .\bar { j }}{\overrightarrow { r } .\bar { k }}+\frac {\overrightarrow { r } .\bar { k }}{\overrightarrow { r } .\bar { i }}=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$
If C is the mid point of AB and P is any point outside AB , then
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$$\overrightarrow{PA}$$ + $$\overrightarrow{PB}$$ + $$\overrightarrow{PC}$$ = 0
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$$\overrightarrow{PA}$$ + $$\overrightarrow{PB}$$ + $$2\overrightarrow{PC}$$ = $$\overrightarrow{0}$$
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$$\overrightarrow{PA}$$ + $$\overrightarrow{PB}$$ = $$\overrightarrow{PC}$$
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$$\overrightarrow{PA}$$ + $$\overrightarrow{PB}$$ = $$2\overrightarrow{PC}$$
A particle in a plane from A to E along the shown path. It is given that AB=BC=DE=10 metre. Then the magnitude of net displacement of particle is :
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10 m
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15 m
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5 m
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20 m
Let $$a=\displaystyle\sum_{i < j}\left(\dfrac{1}{^{n}C_i}+\dfrac{1}{^{n}C_j}\right)$$ and $$b=\displaystyle\sum_{i < j}\left(\dfrac{i}{^{n}C_i}+\dfrac{j}{^{n}C_j}\right)$$, then?
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$$b=(n-1)a$$
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$$b=(n+1)a$$
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$$b=\dfrac{n}{2}a$$
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$$b=na$$
In parallelogram ABCD, $$|\overline { AB } |=a,|\overline { AD } |=b$$ and $$|\overline { AC } |=c$$ then $$\overline { DB } ,\overline { AB } $$ has the value
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$$\frac {3a^2+b^2-c^2}{2}$$
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$$\frac {3b^2+c^2-a^2}{2}$$
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$$\frac {3c^2+b^2-a^2}{2}$$
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$$\frac {a^2+b^2+c^2}{2}$$
If $$ \sum_{i=1}^{n} \vec{a}_{i}=\vec{0} $$ where $$ \left|\vec{a}_{i}\right|=1 \forall i, $$ then the value of $$ \sum_{1 \leq i<j \leq n} \vec{a}_{i} \cdot \vec{a}_{j} $$ is
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$$-n/2$$
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$$-n$$
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$$n/2$$
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$$n$$
If the vectors $$\vec{a}$$ , $$\vec{b}$$ , $$\vec{c}$$ satisfying $$\vec{a}$$ + $$\vec{b}$$ + 2$$\vec{c}$$ = 0 . If $$\left | \vec{a} \right |$$ = 1 , $$\left | \vec{b} \right |$$ = 4 , $$\left | \vec{c} \right |$$ = 2 , then $$\vec{a}.\vec{b}$$ + $$\vec{b}.\vec{c}$$ + $$\vec{c}.\vec{a}$$ =
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$$-\dfrac{7}{2}$$
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$$-\dfrac{17}{2}$$
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$$\dfrac{17}{2}$$
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$$\dfrac{7}{2}$$
The projection of the join of the point (3, 4, 2), (5, 1, 8) on the line whose d.c.'s are $$\left( \dfrac { 2 }{ 7 } ,-\dfrac { 3 }{ 7 } ,\dfrac { 6 }{ 7 } \right) $$ is
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0%
7
0%
$$\left( \dfrac { 46 }{ 13 } \right) $$
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$$\left( \dfrac { 42 }{ 13 } \right) $$
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$$\left( \dfrac { 38 }{ 13 } \right) $$
If $$a$$,$$b$$ and care three mutually perpendicular vectors, then the projection of the vector $$ \left|\frac{a}{|a|}+m \frac{b}{|b|}+n \frac{(a \times b)}{|a \times b|}\right. $$ along the angle bisector of the vector $$a$$ and $$b$$ is
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$$ \frac{l^{2}+m^{2}}{\sqrt{l^{2}+m^{2}-n^{2}}} $$
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$$ \sqrt{1^{2}+m^{2}+n^{2}} $$
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$$ \frac{\sqrt{1^{2}+m^{2}}}{\sqrt{l^{2}-m^{2}+n^{2}}} $$
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$$ \frac{1+m}{\sqrt{2}} $$
Let OAB be a regular triangle with side unity (o being otogin). Also M, N are the points of intersection of AB, M being closer to A and N closer to B. Position vectors of A, B, M and N are $$\vec { a } ,\vec { b } ,\vec { m } $$ and $$\vec { n } $$ respectively. Which of the following hold (s) good ?
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$$\vec { m } =x\vec { a } +y\vec { b } \Rightarrow \dfrac { 2 }{ 3 } $$ and $$y=\dfrac { 1 }{ 3 } $$
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$$\vec { m } =x\vec { a } +y\vec { b } \Rightarrow x=\dfrac { 5 }{ 6 } $$ and $$y=\dfrac { 1 }{ 6 } $$
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$$\vec { m } .\vec { n } $$ equals $$\dfrac { 13 }{ 18 } $$
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$$\vec { m } .\vec { n } $$ equals $$\dfrac { 15 }{ 18 } $$
The position vector of A is $$2\vec { i } +3\vec { j } +4\vec { k } $$$$\vec { AB } =5\vec { i } +7\vec { j } +6\vec { k } $$, then the position vector of B is
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$$-7\vec { i } -10\vec { j } -10\vec { k } $$
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$$7\vec { i } -10\vec { j } +10\vec { k } $$
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$$7\vec { i } +10\vec { j } -10\vec { k } $$
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$$7\vec { i } +10\vec { j } +10\vec { k } $$
The position vector of a point lying on the joining the points whose position vectors are $$\overline i + \overline j -\overline k$$ and $$\overline i - \overline j +\overline k$$ is
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$$\overline j$$
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$$\overline i$$
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$$\overline k$$
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$$\overline 0$$
Area of diagonals is, ..., where diagonals
are
$$a = 2 \hat {i} - 3 \hat { j } + 5 \hat { k } $$, and $$b = - \hat { i} + \hat { j} + \hat { k }$$
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$$\sqrt { 21.5 }$$
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$$\sqrt { 31.5 }$$
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$$\sqrt { 28.5 }$$
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$$\sqrt{ 38.5 }$$
If $$\bar { a } ,\bar { b } ,\bar { c } $$ are position vectors of the points A,B,C respectively such that $$9\bar { a } -7\bar { b } -2\bar { c } =\bar { 0 } $$ then point B divides AC in the ratio.....
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0%
Internally 7:2
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Externally 9:2
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Internally 9:7
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Externally 2:7
A vector $$A=\overrightarrow { l } =\overrightarrow { xj } =3\overrightarrow { k } $$ is rotated through an angle and is also doubled in magnitude resulting in$$\overrightarrow { B } =4\overrightarrow { l } +\left( 4x-2 \right) \overrightarrow { j } +2\overrightarrow { k } $$. An acceptable value of x is
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0%
1
0%
2
0%
3
0%
4/3
A stone projected vertically upwards raises 's' feets in 't' seconds where $$_{ S }=112t-{ 16t }^{ 2 }$$ then the maximum height it reached is
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195 ft
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194 ft
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196 ft
0%
216 ft
Explanation
$$s=112 t-16 t^{2}$$
for maximum height
$$\dfrac{d s}{d t}=0$$
$$\dfrac{d\left(112 t-16 t^{2}\right)}{d t}=0$$
$$\dfrac{d(112 t)}{d t}-\dfrac{d\left(16 t^{2}\right)}{d t}=0$$
$$112-16(2) t^{2-1}=0$$
$$112-32 t=0$$
$$32 t=112$$
$$t=\dfrac{112}{32}=\dfrac{28}{8}=\frac{7}{2}$$
$$\therefore$$ maximum height
$$=112\left(\dfrac{7}{2}\right)-16\left(\dfrac{7}{2}\right)^{2}$$
$$=56 \times 7-4 \times 49$$
$$=392-196$$
$$=196 \mathrm{ft}$$
Given $$\overline { a } = x \hat { i } + y \hat { j } + 2 \hat { k } , \overline { b } = \hat { i } - \hat { j } + \hat { k } , \overline { c } = \hat { i } + 2 \hat { j } ;( \overline { a } \hat { } \overline { b } ) = \pi / 2 , \overline { a } .\overline { c } = 4$$ then
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0%
$$[ \overline { a } \overline { b } \overline { c } ] ^ { 2 } = \left| \overline { a } \right| $$
0%
$$[ \overline { a } \overline { b } \overline { c } ]= \left| \overline { a } \right| $$
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$$[ \overline { a } \overline { b } \overline { c } ]= 0$$
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none of these
If $$ \vec { a } $$ and $$ \vec { b } $$ are vectors such that $$ | \vec { a } + \vec { b } | = \sqrt { 29 } $$ and $$ \vec { a } \times ( 2 \hat { i } + 3 \hat { j } + 4 \hat { k } ) = ( 2 \hat { i } + 3 \hat { j } + 4 \hat { k } ) \times \vec { b } , $$ then
a possible value of $$ ( \vec { a } + \vec { b } ) \cdot ( - 7 \hat { i } + 2 \hat { j } + 3 \hat { k } ) $$ is
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0%
0
0%
3
0%
4
0%
8
For any three $$\vec { a } , \vec { b } , \vec { c } ( \vec { a } - \vec { b } ) ( \vec { b } - \vec { c } ) \times ( \vec { c } - \vec { a } )$$ is equal to
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0%
$$\vec { b } \cdot ( \vec { c } \times \vec { a } )$$
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$$2 \vec { a } \cdot ( \vec { b } \times \vec { c } )$$
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$$0$$
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none of these
The point D,E,F divide BC, CA and Ab of the triangle ABC in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively and the point K divides AB in the ratio 1 : 3 then $$(\overrightarrow { AD } + \overrightarrow { BE } +\overrightarrow { CF }) : \overrightarrow { CK }$$ is equal to
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0%
5 : 2
0%
2:5
0%
1:1
0%
none of these
$$\left( \vec { r } .\vec { i } \right) \left( \vec { r } \times \vec { i } \right) +\left( \vec { r } .\vec { j } \right) \left( \vec { r } \times \vec { j } + \right) \left( \vec { r } .\vec { k } \right) \left( \vec { r } \times \vec { k } \right) $$is equal to
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$$3\vec { r } $$
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$$\vec { r } $$
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$$\vec { 0 } $$
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None of these
If u = $$\hat{j} + 4 \hat{K}, V = \hat{i} - 3 \hat{K} w = cos \theta \hat{i} + sin \theta \hat{j}$$ are vectors in 3- dimensional space, then the maximum possible value of $$|u \times v.w|$$ is
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0%
$$\sqrt{13}$$
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$$\sqrt{14}$$
0%
5
0%
7
If u = $$\hat{j} + 4\hat{k}, V = \hat{i} =- 3K and W = cos \theta i + sin \theta \hat{i}$$ are vectors in 3-dimension space, then the maximum possible value of $$|u \times v. w|$$ is
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0%
$$\sqrt 13$$
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$$\sqrt 14$$
0%
5
0%
7
If $$\hat{i}\times (\vec{a}\times \hat{i})+\hat{j}\times (\vec{a}\times \hat{j})+\hat{k}\times (\vec{a}\times \hat{k})=.....\left\{(\vec{a}.\hat{i})\hat{i}+(\vec{a}.\hat{j})\hat{j}+(\vec{a}.\hat{k})\hat{k}\right\}$$
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0%
$$-1$$
0%
$$0$$
0%
$$2$$
0%
$$None\ of\ these$$
The magnitude of two vectors which can be represented in the form i+j+(2x)k is $$\sqrt{18}$$. Then the unit vector that is perpendicular to these two vectors is
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$$\frac{-i+j}{\sqrt{2}}$$
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$$\frac{i-j}{8\sqrt{2}}$$
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$$\frac{-i+j}{8}$$
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$$\frac{-i+j}{2\sqrt{2}}$$
The length of vector $$ \overrightarrow{A G} $$ is
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$$\sqrt{17}$$
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$$\sqrt{51} / 3$$
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$$3 / \sqrt{6}$$
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$$\sqrt{59} / 4$$
Explanation
Point $$ G $$ is $$ \left(\dfrac{4}{3}, \dfrac{1}{3}, \dfrac{8}{3}\right) \cdot $$ Therefore
$$\begin{array}{l}|\overrightarrow{A G}|^{2}=\left(\dfrac{5}{3}\right)^{2}+\dfrac{1}{9}+\left(\dfrac{5}{3}\right)^{2}=\dfrac{51}{9} \\\Rightarrow|\overrightarrow{A G}|=\dfrac{\sqrt{51}}{3} \\\overrightarrow{A B}=-4 \hat{i}+4 \hat{j}+0 \hat{k} \\\overrightarrow{A C}=2 \hat{i}+2 \hat{j}+2 \hat{k}\end{array}$$
$$ \therefore \overrightarrow{A B} \times \overrightarrow{A C}=-8\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 1 & 1\end{array}\right| $$
$$ =8(\hat{i}+\hat{j}-2 \hat{k}) $$
Area of $$ \Delta A B C=\dfrac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|=4 \sqrt{6} $$
$$ \overrightarrow{A D}=-3 \hat{i}-5 \hat{j}+3 \hat{k} $$
The length of the perpendicular from the vertex $$ D $$ on the opposite face
$$ =\mid $$ projection of $$ \overrightarrow{A D} $$ on $$ \overrightarrow{A B} \times \overrightarrow{A C} \mid $$
$$\begin{array}{l}=\left|\dfrac{(-3 \hat{i}-5 \hat{j}+3 \hat{k})(\hat{i}+\hat{j}-2 \hat{k})}{\sqrt{6}}\right| \\ \\=\left|\dfrac{-3-5-6}{\sqrt{6}}\right|=\dfrac{14}{\sqrt{6}}\end{array}$$
For non-zero vectors $$ \vec{a}, \vec{b} $$ and $$ \vec{c},|(\vec{a} \times \vec{b}) \cdot \vec{c}|=|\vec{a}||\vec{b}||\vec{c}| $$ holds if and only if
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$$ \vec{a} \cdot \vec{b}=0, \vec{b} \cdot \vec{c}=0 $$
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$$\vec{b} \cdot \vec{c}=0, \vec{c} \cdot \vec{a}=0 $$
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$$ \vec{c} \cdot \vec{a}=0, \vec{a} \cdot \vec{b}=0 $$
0%
$$\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$$
Explanation
$$\textbf{Step 1: Simplifying given expression}$$
$$\text{Given that }$$
$$|(\vec a \times \vec b).\vec c|=|\vec a||\vec b||\vec c|$$
$$\implies ||\vec a||\vec b|sin\theta \hat n.\vec c|=|\vec a||\vec b||\vec c|$$
$$\text{Here }\hat n \text{ is the resultant of vector } a\text{ and }b$$
$$\text{and } \hat n.\vec c=|\hat n||\vec c|cos\alpha$$
$$\implies ||\vec a||\vec b|sin\theta |\hat n||\vec c|cos\alpha |=|\vec a||\vec b||\vec c|$$
$$\text{Since }\hat n \text{ is a unit vector }|\hat n|=1$$
$$\textbf{Step 2: Calculating}$$
$$\implies ||\vec a||\vec b||\vec c|sin\theta.cos\alpha|=|\vec a ||\vec b||\vec c|$$
$$\implies |sin\theta ||cos\alpha|=1$$
$$\textbf{Step 3: Calculating result}$$
$$\text{If we assume that }\theta = \dfrac {\pi}2 , \alpha =0$$
$$\implies \vec a \text{ and } \vec b \text{ are perpendicular}$$
$$\implies \vec a.\vec b=0$$
$$\text{As }\alpha=0, \vec c \text{ is along the resultant of vector }a\text{ and }b$$
$$\implies \vec c \text{ is perpendicular to }\vec a \text { and }\vec b$$
$$\implies \vec c.\vec a=0 \text{ and } \vec c.\vec b=0$$
$$\textbf{Hence, }\mathbf{\vec a.\vec b= \vec a.\vec c= \vec b.\vec c=0}$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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