Vector Algebra

If \left| {\matrix{ a & {{a^2}} & {1 + {a^3}} \cr b & {{b^2}} & {1 + {b^3}} \cr c & {{c^2}} & {1 + {c^3}} \cr } } \right| = 0 and vectors $\left( {1,a,{a^2}} \right),\,\,$ $\left( {1,b,{b^2}} \right)$ and $\left( {1,c,{c^2}} \right)\,$ are non-coplanar, then the product $abc$ equals :

Consider points $A, B, C$ and $D$ with position vectors $7\widehat i - 4\widehat j + 7\widehat k,\widehat i - 6\widehat j + 10\widehat k, - \widehat i - 3\widehat j + 4\widehat k$ and $5\widehat i - \widehat j + 5\widehat k$ respectively. Then $ABCD$ is a :

The vectors $\overrightarrow {AB} = 3\widehat i + 4\widehat k\,\,\& \,\,\overrightarrow {AC} = 5\widehat i - 2\widehat j + 4\widehat k$ are the sides of triangle $ABC.$ The length of the median through $A$ is :

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

If $C$ is the mid point of $AB$ and $P$ is any point outside $AB,$ then :

Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a\widehat i + a\widehat j + c\widehat k,\,\,\widehat i + \widehat k$ and $c\widehat i + c\widehat j + b\widehat k$ lie in a plane, then $c$ is :

The non-zero vectors are ${\overrightarrow a ,\overrightarrow b }$ and ${\overrightarrow c }$ are related by ${\overrightarrow a = 8\overrightarrow b }$ and ${\overrightarrow c = - 7\overrightarrow b \,\,.}$ Then the angle between ${\overrightarrow a }$ and ${\overrightarrow c }$ is :

Let $\overrightarrow a$, $\overrightarrow b$, $\overrightarrow c$ be three non-zero vectors which are pairwise non-collinear. If $\overrightarrow a+3 \overrightarrow b$ is collinear with $\overrightarrow c$ and $\overrightarrow b+2 \overrightarrow c$ is collinear with $\overrightarrow a$, then $\overrightarrow a+\overrightarrow b+6 \overrightarrow c$ is :

If the vectors $\overrightarrow {AB} = 3\widehat i + 4\widehat k$ and $\overrightarrow {AC} = 5\widehat i - 2\widehat j + 4\widehat k$ are the sides of a triangle $ABC,$ then the length of the median through $A$ is :

Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $\overrightarrow a ,\overrightarrow b ,\overrightarrow c$ and ${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$ respectively, then the position vector of the orthocentre of this triangle, is :

If the position vectors of the vertices A, B and C of a $\Delta$ ABC are respectively $4\widehat i + 7\widehat j + 8\widehat k,$    $2\widehat i + 3\widehat j + 4\widehat k,$ and $2\widehat i + 5\widehat j + 7\widehat k,$ then the position vectors of the point, where the bisector of $\angle$A meets BC is :

If $\overrightarrow \alpha$ = $\left( {\lambda - 2} \right)\overrightarrow a + \overrightarrow b$  and  $\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow a + 3\overrightarrow b$ be two given vectors $\overrightarrow a$ and $\overrightarrow b$ are non-collinear. The value of $\lambda$ for which vectors $\overrightarrow \alpha$ and $\overrightarrow \beta$ are collinear, is -

The lines $\overrightarrow r = \left( {\widehat i - \widehat j} \right) + l\left( {2\widehat i + \widehat k} \right)$ and $\overrightarrow r = \left( {2\widehat i - \widehat j} \right) + m\left( {\widehat i + \widehat j + \widehat k} \right)$

If vectors $\overrightarrow {{a_1}} = x\widehat i - \widehat j + \widehat k$ and $\overrightarrow {{a_2}} = \widehat i + y\widehat j + z\widehat k$ are collinear, then a possible unit vector parallel to the vector $x\widehat i + y\widehat j + z\widehat k$ is :

Let a vector $\alpha \widehat i + \beta \widehat j$ be obtained by rotating the vector $\sqrt 3 \widehat i + \widehat j$ by an angle 45$^\circ$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices ($\alpha$, $\beta$), (0, $\beta$) and (0, 0) is equal to :

A vector $\overrightarrow a$ has components 3p 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 new system, $\overrightarrow a$ has components p + 1 and $\sqrt {10}$, then the value of p is equal to :

Let a, b and c be distinct positive numbers. If the vectors $a\widehat i + a\widehat j + c\widehat k,\widehat i+\widehat k$ and $c\widehat i + c\widehat j + b\widehat k$ are co-planar, then c is equal to :

Let A, B, C be three points whose position vectors respectively are $\overrightarrow a = \widehat i + 4\widehat j + 3\widehat k$ $\overrightarrow b = 2\widehat i + \alpha \widehat j + 4\widehat k,\,\alpha \in R$ $\overrightarrow c = 3\widehat i - 2\widehat j + 5\widehat k$ If $\alpha$ is the smallest positive integer for which $\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c$ are noncollinear, then the length of the median, in $\Delta$ABC, through A is :

Let the vectors $\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}, \vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$ and $\vec{c}=t \hat{i}-t \hat{j}+\hat{k}, t \in \mathbf{R}$ be such that for $\alpha, \beta, \gamma \in \mathbf{R}, \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0} \Rightarrow \alpha=\beta=\gamma=0$. Then, the set of all values of $t$ is :