CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 3 - MCQExams.com

find the coordinate of the tip of the position vector which is equivalent to $$ \vec{AB}$$, where the coordinates of A and B are (-1, 3) and (-2, 1) respectively.
  • (+1,+2)
  • (+1,-2)
  • (-1,+2)
  • (-1,-2)
If the position vectors of the points $$A(3,4),B(5, -6)$$ and $$C(4,-1)$$ are $$ \vec{a}, \vec{b}, \vec{c}$$ respectively, compute $$ \vec{a}+2\vec{b}-3\vec{c}. $$
  • $$ -5\hat{i}-1\hat{j} $$
  • $$ \hat{i}-5\hat{j} $$
  • $$ \hat{i}+5\hat{j} $$
  • none of these
If $$4 \hat{i} + 7 \hat{j} + 8 \hat{k},  2 \hat{i} + 3 \hat{j} + 4 \hat{k}$$ and $$ 2 \hat{i} + 5 \hat{j} + 7 \hat{k}$$ are the position vectors of the vertices $$A, B$$ and $$C$$ respectively, of the triangle $$ABC$$, the position vector of the point where the bisector of angle $$A$$ meets $$BC$$, is
  • $$\frac{2}{3} (-6 \hat{i} - 8 \hat{j} - 6 \hat{k})$$
  • $$\frac{2}{3} (6 \hat{i} + 8 \hat{j} + 6 \hat{k})$$
  • $$\frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})$$
  • $$\frac{1}{3} ( 5 \hat{j} + 12 \hat{k})$$
If vectors $$\overrightarrow{AB} = -3 \hat{j} + 4 \hat{k}$$ and $$\overrightarrow{AC} = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$$ are the sides of a $$\Delta ABC$$, then the length of the median through $$A$$ is
  • $$\sqrt{14}$$
  • $$\sqrt{18}$$
  • $$\sqrt{29}$$
  • $$5$$
Let $$ABC$$ be a triangle, the position vectors of whose vertices are respectively $$\hat{i} + 2 \hat{j} + 4 \hat{k}, -2 \hat{i} + 2 \hat{j} + \hat{k}$$ and $$2 \hat{i} + 4 \hat{j} + 3 \hat{k}$$. Then $$\Delta ABC$$ is 
  • isosceles
  • equilateral
  • right angled
  • none of these
If $$\vec a$$ is parallel to $$\vec b \times \vec c$$, then $$(\vec a \times \vec b) \cdot (\vec a \times \vec c)$$ is equal to
  • $$\mid \vec a \mid^2 (\vec b \cdot \vec c)$$
  • $$\mid \vec b \mid^2 (\vec a \cdot \vec c)$$
  • $$\mid \vec c \mid^2 (\vec a \cdot \vec b)$$
  • none of these
If $$\mid \vec a \mid = 2$$ and $$\mid \vec b \mid = 3$$ and $$\vec a \cdot \vec d = 0$$, then $$(\vec a \times (\vec a \times (\vec a \times (\vec a \times \vec b ))))$$ is equal to
  • $$48 \hat b$$
  • $$- 48 \hat b$$
  • $$48 \hat a $$
  • $$48 \hat a$$
Figure shows $$ABCDEF$$ as a regular hexagon. What is the value of 
$$\vec{AB}+\vec{AC}+\vec{AD}+\vec{AE}+\vec{AF}$$
1811900_932009e245e34da694a5d16a66b21bd8.png
  • $$\vec{AO}$$
  • $$2\vec{AO}$$
  • $$4\vec{AO}$$
  • $$6\vec{AO}$$
Let $$\mathrm{A}\mathrm{B}\mathrm{C}$$ be a triangle and let $$\mathrm{D},\mathrm{E}$$ be the midpoints of the sides $$\mathrm{A}\mathrm{B},\mathrm{A}\mathrm{C}$$ respectively,then $$\hat { BE } +\hat { DC } =$$
  • $$\hat { BC } $$
  • $${ \dfrac { 1 }{ 2 } }\hat { BC } $$
  • $$ { \dfrac { 3 }{ 2 } \hat { BC } } $$
  • $$\dfrac { 3 }{ 4 } \hat { BC } $$
Let  $$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$$  and  $$\vec{c}=\hat{i}-\hat{j}-\hat{k}$$  be three vectors. A vector  $$\vec{v}$$  in the plane of   $$\vec{a}$$ and $$\vec{b}$$ , whose projection on  $$\vec{c}$$  is $$\displaystyle \dfrac{1}{\sqrt{3}}$$ , is given by $$;$$
  • $$\hat{i}-3\hat{j}+3\hat{k}$$
  • $$-3\hat{i}-3\hat{j}-\hat{k}$$
  • $$3\hat{i}-\hat{j}+3\hat{k}$$
  • $$\hat{i}+3\hat{j}-3\hat{k}$$
The position vectors of $$A,B,C$$ are $$\overline{i}+\overline{j}+\overline{k},\ 4\overline{i}+\overline{j}+\overline{k},\ 4\overline{i}+5\overline{j}+\overline{k}$$ . Then the position vector of the circumcentre of the triangle $$ABC$$ is
  • $$3\hat{i}+2\hat{j}+\hat{k}$$
  • $$\displaystyle \frac{1}{2}(6\hat{i}+\hat{j}+\hat{k})$$
  • $$\displaystyle \frac{1}{2}(5\hat{i}+6\hat{j}+2\hat{k})$$
  • $${\dfrac{1}{2}(9\overline{i}}+7\overline{j}+3\overline{k})$$
The vector sum of $$N$$ coplanar forces, having magnitude of $$F$$, when each force is making an angle of $$\displaystyle \frac{2\pi}{N}$$ with that preceding it, is:
  • $$F$$
  • $$\displaystyle \frac{NF}{2}$$
  • $$NF$$
  • $$0$$
If the vectors  $$\overline{a}=3\overline{i}+\overline{j}-2\overline{k}$$,$$\overline{b}=-\overline{i}+3\overline{j}+4\overline{k},\ \overline{c}=4\overline{i}-2\overline{j}-6\overline{k}$$  form the sides of the triangle then length of the median bisecting the vector $${c}$$ is
  • $$\sqrt{12}$$ units
  • $$\sqrt{6}$$ units
  • $$2\sqrt{6}$$ units
  • $$2\sqrt{3}$$ units
 If $$\mathrm{O}$$ is the circumcentre and $$\mathrm{O}^{'}$$ is the orthocentre of a triangle $$\mathrm{A}\mathrm{B}\mathrm{C}$$ and if $$\mathrm{A}\mathrm{P}$$ is the circumdiameter then
$$\vec{\mathrm{A}\mathrm{O}}+\vec{\mathrm{O}^{'}\mathrm{B}}+\vec{\mathrm{O}^{'}\mathrm{C}}=$$
  • $$\vec{\mathrm{O}\mathrm{A}}$$
  • $$\vec{\mathrm{O}^{'}\mathrm{A}}$$
  • $$\vec{\mathrm{A}\mathrm{P}}$$
  • $$\vec{\mathrm{A}\mathrm{O}}$$
Five equal forces each of 20N are acting at a point in the same plane. If the angles between them are same, then the resultant of these forces is:
  • 0
  • 40N
  • 20N
  • $$20\sqrt{2}$$
If there are 11 vectors each having a magnitude equal to $$| \vec{p} |$$ and if each side of polygon subtends an angle $$30^{0}$$ at the centre of the polygon. Then the resultant is 
  • $$\vec{p}$$
  • 0
  • $$\frac{\vec{p}}{2}$$
  • 2$$\vec{p}$$
$$\mathrm{If}$$ $$\vec{AD},\ \vec{BE},\ \vec{CF}$$ are medians of an equilateral triangle $$\mathrm{A}\mathrm{B}\mathrm{C}$$, then $$\vec{AD}+\vec{BE}+\vec{CF}$$ equals to 
  • $$\vec{AB}+\vec{BC}+\vec{CA}$$
  • a zero vector
  • both $$1$$ and $$2$$
  • $$2\vec{AF}+3\vec{BF}$$
If $$\overrightarrow { a } ,\overrightarrow { b } $$ and $$\overrightarrow { c } $$ are three non-zero vectors such that $$\overrightarrow { a } .\overrightarrow { b } =\overrightarrow { a } .\overrightarrow { c } ,$$ then
  • $$\overrightarrow { b } =\overrightarrow { c } $$
  • $$\overrightarrow { a } \bot \overrightarrow { b } ,\overrightarrow { c } $$
  • $$\overrightarrow { a } \bot \overrightarrow { b } -\overrightarrow { c } $$
  • either $$\overrightarrow { a } \bot (\overrightarrow { b } -\overrightarrow { c }) $$ or $$\overrightarrow { b } =\overrightarrow { c } $$
Let $$\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$$ be a parallelogram and let $$\mathrm{L}$$ and $$\mathrm{M}$$ be the midpoints of the sides $${ BC } $$ and $$ { CD }$$  respectively. Then  $$\vec { AL } +\vec { AM } =$$
  • $$\vec { AC } $$
  • $${ \dfrac { 2 }{ 3 } }\vec { AC } $$
  • $$\dfrac { 3 }{ 2 } \vec { AC } $$
  • $$2\vec { AC } $$
Let $$\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$$ be a trapezium and let $$\mathrm{P},\mathrm{Q}$$ be the midpoints of the nonparallel sides $$\mathrm{A}\mathrm{D},\ \mathrm{B}\mathrm{C}$$ respectively. Then $$\vec { PQ } $$
  • $$\vec { AB } +\vec { DC }$$
  • $${ \dfrac { 1 }{ 4 } }(\vec { AB } +\vec { BC } )$$
  • $$\dfrac { 1 }{ 2 } (\vec { AB } +\vec { DC } )$$
  • $$\dfrac { 1 }{ 2 } (\vec { AB } +\vec { BC } )$$
$${A}{B}{C}{D}$$ is a quadrilateral, $$E$$ is the point of intersection of the line joining the middle points of the opposite sides. lf $$O$$ is any point, then $$\hat { OA } +\hat { OB } +\hat { OC } +\hat { OD } =$$
  • $$4\hat { OE } $$
  • $$3\hat { OE } $$
  • $$2\hat { OE } $$
  • $$\hat { OE } $$
If $$\overline{DA}=\overline{a};\overline{AB}=\overline{b}$$ and $$\overline{CB}=k\overline{a}$$ where $$k>0$$ and $$x, y$$ are the midpoints of $$DB$$ and $$AC$$ respectively such that $$|\overline{a}|=17$$ and $$|\overline{XY}|=4$$, then $$\mathrm{k}=$$
  • $$\displaystyle \dfrac{8}{17}$$
  • $$\displaystyle \dfrac{9}{17}$$
  • $$\displaystyle \dfrac{11}{17}$$
  • $$\displaystyle \dfrac{4}{17}$$
If the vectors $$4\hat{i}-7\hat{j}-2\hat{k},\: \hat{i}+5\hat{j}-3\hat{k},\: 3\hat{i}-\lambda\hat{j}+\hat{k}$$ form a triangle then $$\lambda=$$
  • $$6$$
  • $$-6$$
  • $$12$$
  • $$-1$$
The incentre of the triangle formed by the points $$ { \hat { i }  }+{ \hat { j }  }+{ \hat { k }  },$$ $$ 4{ \hat { i }  }+{ \hat { j }  }+{ \hat { k }  }$$, and $$4{ \hat { i }  }+5{ \hat { j }  }+\hat { k }$$  is
  • $$\dfrac { { \hat { i } }+{ \hat { j } }+{ \hat { k } } }{ 3 } $$
  • $${ \hat { i } }+2{ \hat { j } }+3\hat { k } $$
  • $$3{ \hat { i } }+2{ \hat { j } }+\hat { k } $$
  • $${ \hat { i } }+{ \hat { j } }+{ \hat { k } }$$
lf $$4{\vec{i}}+7\vec{j}+8\vec{k} , 2\vec{i}+3\vec{j}+4\vec{k}$$ and $$2\vec{i}+5\vec{j}+7\vec{k}$$ are the position vectors of the vertices $$\mathrm{A},\mathrm{B}$$ and $$\mathrm{C}$$ of $$ \triangle \mathrm{A}\mathrm{B}\mathrm{C}$$, the position vector of $$D$$ the point where the bisector of $$\angle A$$ meets $$\mathrm{B}\mathrm{C}$$ is
  • $${\dfrac{2}{3}}(-6\hat {i}-8\hat {j}-6\hat {k})$$
  • $${\dfrac{2}{3}}(6\hat {i}+8\hat{j}+6\hat{k})$$
  • $${\dfrac{1}{3}}(6\hat{i}+13\hat{j}+18\hat{k})$$
  • $$2(\hat{i}+\hat{j}+\hat{k})$$
$$\hat { AB } =-3{ \hat { i }  }+4{ \hat { k }  }$$ and $$\hat { BC } =-\overline { i } -2\overline { k } $$ are the sides of the triangle $$ ABC$$ then the length of the median $$AM$$ is
  • $$\sqrt{\dfrac{25}{2}}$$
  • $$\sqrt{\dfrac{45}{2}}$$
  • $$\displaystyle \dfrac{\sqrt{65}}{2}$$
  • $$\displaystyle \dfrac{\sqrt{85}}{2}$$
lf $$A=(-3,2,5), B=(-3,4,5)$$ and $$C=(-3,4,7)$$ are the position vectors of vertices of $$\Delta ABC$$ then its circumcentre is
  • $$(-3,3,5)$$
  • $$(-3,3,6)$$
  • $$(-3,4,6)$$
  • $$(-3,4,7)$$
lf $$\vec{a}$$ and $$\vec{b}$$ are two non-parallel unit vectors and the vector $$\alpha\vec{a}+\vec{b}$$ bisects the internal angle between $$\vec{a}$$ and $$\vec{b}$$, then $$\alpha$$ is
  • $$1$$
  • $$\dfrac{1}{2}$$
  • $$2$$
  • $$5$$
Two forces act at the vertex $$\mathrm{A}$$ of quadrilateral $$\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$$ represented by $$\overline{AB},\ \overline{AD}$$ and two at $$\mathrm{C}$$ represented by $$\overline{CD}$$ and $$\overline{CB}$$. If $$\mathrm{E},\ \mathrm{F}$$ are mid points of $$\overline{AC}$$ and $$\overline{BD}$$ respectively, then their resultant is
  • $$\overline{EF}$$
  • $$2\overline{EF}$$
  • $${\dfrac{3}{2}\overline{EF}}$$
  • $$4\overline{EF}$$
If point $$O$$ is the centre of a circle circumscribed about a triangle $$ABC$$. Then $$\overline{OA}\sin 2A+\overline{OB}\sin2B+\overline{OC}\sin 2C=$$
  • $$(\overline{OA}+\overline{OB}+\overline{OC})\sin 2A$$
  • $$(\overline{OA}+\overline{OB}+\overline{OC})\cos 2A$$
  • $$\overline{0}$$
  • $$(\overline{OA}+\overline{OB}+\overline{OC})\tan 2A$$
Let $$G$$ and $$G^{1}$$ be the centroids of the triangles $$ABC$$ and $$A^{1}B^{1}C^{1}$$ respectively, then $$AA^{1}+BB^{1}+CC^{1}$$ is equal to
  • $$2GG^{1}$$
  • $$3G^{1}G$$
  • $$3GG^{1}$$
  • $$\displaystyle \dfrac{3}{2}GG^{1}$$
The ratio in which $$\overline{i}+2\overline{j}+3\overline{k}$$ divides the join $$\mathrm{o}\mathrm{f}-2\overline{i}+3\overline{j}+5\overline{k}$$ and $$7\overline{i}-\overline{k}$$ is
  • $$-3:2$$
  • $$1:2$$
  • $$2:3$$
  • $$-4:3$$
Orthocentre of an equilateral triangle $$ABC$$ is the origin $$\mathrm{O}$$. If $$A=\overline{a},\ B=\overline{b},\ C=\overline{c}$$ then $$\overline{AB}+2\overline{BC}+3\overline{CA}=$$
  • $$3\overline{c}$$
  • $$3\overline{a}$$
  • $$0$$
  • $$3\overline{b}$$
If the vertices of a $$\Delta ABC$$ are $$A= (1,-1,-3)$$ , $$B= (2, 1, -2)$$ and $$C=(-5,2,-6)$$ then the length of the internal bisector of angle $$A$$ is
  • $$\displaystyle \dfrac{3\sqrt{10}}{2}$$
  • $$\displaystyle \dfrac{3\sqrt{10}}{5}$$
  • $$\displaystyle \dfrac{3\sqrt{10}}{7}$$
  • $$\displaystyle \dfrac{3\sqrt{10}}{4}$$
The condition for the vectors $$\vec{a},\vec{b},\vec{c},\vec{d}$$ to be the sides of a parallelogram taken in order is
  • $$\vec{a}+\vec{b}=\vec{c}+\vec{d}$$
  • $$\vec{a}+\vec{b}=\vec{c}+\vec{d}=0$$
  • $$\vec{a}+\vec{c}=\vec{b}+\vec{d}$$
  • $$\vec{a}+\vec{c}=\vec{b}+\vec{d}=0$$
Let $$A$$ and $$B$$ be points with position vectors $$\overline{a}$$ and $$\overline{b}$$ with respect to origin $$O$$. If the point $$C$$ on $$OA$$ is such that $$2\overline{AC}=\overline{CO}$$, $$\overline{CD}$$ is parallel to $$\overline{OB}$$ and $$|\overline{CD}|=3|\overline{OB}|$$ then $$AD$$ is
  • $$\displaystyle \overline{b}-\dfrac{a}{9}$$
  • $$3\displaystyle \overline{b}-\dfrac{a}{3}$$
  • $$\displaystyle \overline{b}-\dfrac{a}{3}$$
  • $$\displaystyle \overline{b}+\dfrac{a}{3}$$
The adjacent sides of a parallelogram are $$2{ \hat { i }  }+4{ \hat { j }  }-5{ \hat { k }  }$$ and $$ { \hat { i }  }+2{ \hat { j }  }{ + }3{ \hat { k }  }$$ then the unit vector parallel to a diagonal is
  • $$\displaystyle \dfrac{-\overline{i}-2\overline{j}+8\overline{k}}{\sqrt{69}}$$
  • $$\displaystyle \dfrac{3\overline{i}+6\overline{j}-2\overline{k}}{7}$$
  • $$\displaystyle \dfrac{\overline{i}+2\overline{j}-8\overline{k}}{\sqrt{69}}$$
  • All the above
If the diagonals of a parallelogram are $$\overline{i}+5\overline{j}-2\overline{k}$$ and $$-2\overline{i}+\overline{j}+3\overline{k}$$, then the lengths of its sides are
  • $$\displaystyle \dfrac{\sqrt{38}}{2},\ \displaystyle \dfrac{\sqrt{50}}{2}$$
  • $$\displaystyle \dfrac{\sqrt{35}}{2},\dfrac{\sqrt{45}}{2}$$
  • $$\sqrt{38},\ \sqrt{50}$$
  • $$\sqrt{35},\ \sqrt{45}$$
Let $$A=2\hat{i}+4\hat{j}-\hat{k}$$, $$B=4\hat{i}+5\hat{j}{+}\hat{k}$$. If the centroid $$G$$ of the triangle $$ABC$$ is $$3\hat{i}+5\hat{j}-\hat{k}$$, then the position vector of $$C$$ is
  • $$3\hat{i}-6\hat{j}{+}3\hat{k}$$
  • $$3\hat{i}-6\hat{j}-3\hat{k}$$
  • $$3\hat{i}-6\hat{j}{+}2\hat{k}$$
  • $$3\hat{i}+6\hat{j}-3\hat{k}$$
If $$I$$ is the centre of a circle inscribed in a triangle $$ABC$$, then $$|\overline{BC}|\overline{IA}+|\overline{CA}|\overline{IB}+|\overline{AB}|\overline{IC}$$ is
  • $$\overline{0}$$
  • $$\overline{IA}+\overline{IB}+\overline{IC}$$
  • $$\displaystyle \frac{\overline{IA}+\overline{IB}+\overline{IC}}{3}$$
  • $$\displaystyle \frac{\overline{IA}+\overline{IB}+\overline{IC}}{2}$$
If $$\vec{{r}} {\times} \vec{{a}}=\vec{{b}}{\times}\vec{{a}};\ \vec{{r}}{\times}\vec{{b}}=\vec{{a}}{\times}\vec{{b}};\ \vec{a}\neq 0,\vec{b}\neq 0,\vec{a}\neq\lambda\vec{b};\ \vec{a}$$ is not perpendicular to $$\vec{b}$$, then $$\vec{r}=$$
  • $$\vec{{a}}-\vec{{b}}$$
  • $$\vec{{a}}+\vec{{b}}$$
  • $$\vec{{a}}{\times}\vec{{b}}+\vec{{a}}$$
  • $$\vec{{a}}{\times}\vec{{b}}+\vec{{b}}$$
Let $$A({\vec{a}})$$ , $$B({\vec{b}}), C({\vec{c}})$$ be the vertices of the triangle $$ABC$$ and let $$DEF$$ be the mid points of the sides $$BC, CA, AB$$ respectively. If $$P$$ divides the median $$AD$$ in the ratio $$2:1$$ then the position vector of $$P$$ is
  • $$0$$
  • $${\vec{a}}+{\vec{b}}+{\vec{c}}$$
  • $$\displaystyle \dfrac{\vec{{a}}+\vec{{b}}+\vec{{c}}}{3}$$
  • $$\displaystyle \dfrac{2{\vec{a}}+{\vec{b}}+{\vec{c}}}{3}$$
The plane $$2x-3y+z+6=0$$ divides the line segment joining $$(2, 4, 16)$$ and $$(3, 5, -4)$$ in the ratio
  • $$ 4 :5 $$
  • $$ 4:7 $$
  • $$2:1$$
  • $$1:2$$
The position vectors of points $$\vec{A},\vec{B},\vec{C}$$ are respectively $$\vec{a},\vec{b},\vec{c}$$. If $$P$$ divides $$\vec{AB}$$ in the ratio $$3:4$$ and $$Q$$ divides $$\vec{BC}$$ in the ratio $$2:1$$ both externally then $$\vec{PQ}$$ is
  • $$\vec{b}+\vec{c}-\vec{2a}$$
  • $$2(\vec{b}+\vec{c}-\vec{2a})$$
  • $$4\vec{a}-\vec{b}-\vec{c}$$
  • $$\dfrac{-2\vec{a}-\vec{b}-\vec{c}}{2}$$
If the position vectors of $$A,B,C$$ are $$3\hat{i}+\hat{j}-\hat{k}$$, $$\hat{i}-2\hat{j}$$, $$2\hat{i}-\hat{j}{+}3\hat{k}$$ then the position vector of the centroid of the triangle $$ABC$$ is
  • $$2\hat{i}\ + \dfrac{-2}{3}\hat{j}\ +\dfrac{2}{3}\hat{k}$$
  • $$2\hat{i}\ + \dfrac{2}{3}\hat{j}\ +\dfrac{2}{3}\hat{k}$$
  • $$3\hat{i}\ + \dfrac{2}{3}\hat{j}\ +\dfrac{4}{3}\hat{k}$$
  • None of these
Let $$\vec{A}=2\hat{i}+7\hat{j},\vec{B}=\hat{i}+2\hat{j}+4\hat{k},\ \displaystyle \vec{C}=\dfrac{9\hat{i}+30\hat{j}+4\hat{k}}{5}$$. 
The ratio in which $$\vec{C}$$ divides $$\vec{AB}$$ internally is?
  • $$1:4$$
  • $$2:3$$
  • $$3:2$$
  • $$5:1$$
If $$\vec a, \vec b, \vec c, \vec d$$ are the position vectors of the points $$A, B, C, D$$ respectively such that $$3\vec{a}+5\vec{b}-3\vec{c}-5\vec{d}=0$$ then $$AB$$ intersects $$CD$$ in the ratio
  • $$2:3$$
  • $$3:2$$
  • $$3:5$$
  • $$5:3$$
If $$(2, -1, 2)$$ is the centroid of tetrahedron $$OABC$$ and $$G_{1}$$ is the centroid of $$\Delta ABC$$ then $$|\overline{OG}_{1}|=$$
  • $$4$$
  • $$1$$
  • $$\dfrac{9}{2}$$
  • $$\dfrac{3}{2}$$
If $$\overline{a}=\overline{i}+\overline{j}+\overline{k},\overline{b}=2\overline{i}-3\overline{j}+\overline{k}$$, then $$\displaystyle \dfrac{\overline{a}\times\overline{b}}{|\overline{a}\times\overline{b}|}+\dfrac{\overline{b}\times\overline{a}}{|\overline{b}\times\overline{a}|}=$$
  • $$\overline{0}$$
  • $$2\overline{i}+\overline{j}-2\overline{k}$$
  • $$\overline{i}+\overline{j}+2\overline{k}$$
  • $$\overline{i}+2\overline{j}-\overline{k}$$
The position vectors of $$A, B , C, D$$ are $$\overline{a},\overline{b},\ \overline{c},\overline{d}$$ respectively and $$|\overline{a}-\overline{d}|=|\overline{b}-\overline{d}|=|\overline{c}-\overline{d}|$$, then for the triangle $$ABC$$, $$D$$ is
  • Ortho centre
  • Centroid
  • Circumcentre
  • Incentre
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