Vector Algebra

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 :

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 -

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 :

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 :

Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that $${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$$. Then $${{Area(\Delta PQR)} \over {Area(\Delta ABC)}}$$ is equal to :