Test for Collinearity of Points

Q.

If the coordinates of two points are $P$$\left(-2,3\right)$ and $Q$$\left(-3,5\right)$ then find abscissa of $P-$ abscissa $Q$

Q.

In following figure, the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of Z =x + 2y.

Q. The vectors $2\hat{i}+3\hat{j},5\hat{i}+6\hat{j}$ and $8\hat{i}+\alpha \hat{j}$ have their initial points at $(1,1)$. The value of $\alpha$ so that the vectors terminate on one straight line is

1. $9$
2. $3$
3. $6$
4. $0$

Q. A point $(p,q,r)$ lies on the plane $\overrightarrow{r}\cdot (\hat{i}+2\hat{j}+\hat{k})=4$. The value of $q$ such that the vector $\overrightarrow{a}=p\hat{i}+q\hat{j}+r\hat{k}$ satisfies the relation $\hat{j}\times (\hat{j}\times \overrightarrow{a})=\overrightarrow{0},$ is

Q. $P(1,0,1),Q(2,0,-3),R(-1,2,0)$ and $S(3,-2,-1)$, then find the projection length of $\overrightarrow{PQ}$ on $\overrightarrow{RS}$

Q. The points A, B and C with position vectors $-2\hat{i}+3\hat{j}+5\hat{k},\hat{i}+2\hat{j}+3\hat{k}and7\hat{i}-\hat{k}$ respectively are non collinear.

1. False
2. True

Q. Show that the vectors $2\hat{i}-3\hat{j}+4\hat{k}$ and $-4\hat{i}+6\hat{j}-8\hat{k}$ are collinear

Q. A point $(p,q,r)$ lies on the plane $\overrightarrow{r}\cdot (\hat{i}+2\hat{j}+\hat{k})=4$. The value of $q$ such that the vector $\overrightarrow{a}=p\hat{i}+q\hat{j}+r\hat{k}$ satisfies the relation $\hat{j}\times (\hat{j}\times \overrightarrow{a})=\overrightarrow{0},$ is

Q. If $A=(5,6,7,8,9),B=\{x:3<x<8\text{and}x\in W\}$ and $C=\{x:x\le 5\text{and}x\in N\}$. Find number of elements in A $\cap$ B and (A $\cap$ B) $\cap$ C.

Q. If $\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{BO}+\overrightarrow{OC}$, then

1. A, B and C are collinear
2. A, B and C are not collinear
3. Nothing can be said about Collinearity from this information
4. $\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}$ form a triangle

Q. If $\overrightarrow{A}=(\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c})$
$\overrightarrow{B}=(-2\overrightarrow{a}+3\overrightarrow{b}+2\overrightarrow{c})$
$\overrightarrow{C}=(-8\overrightarrow{a}+13\overrightarrow{b})$
Then $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are collinear

1. True
2. False

Q. If the position vectors of three points are $\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c},2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c},-7\overrightarrow{b}+10\overrightarrow{c}$, then the three points are

1. collinear
2. non – coplanar
3. non – collinear
4. None of these

Q. If $\overrightarrow{A}=(\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c})$
$\overrightarrow{B}=(-2\overrightarrow{a}+3\overrightarrow{b}+2\overrightarrow{c})$
$\overrightarrow{C}=(-8\overrightarrow{a}+13\overrightarrow{b})$
Then $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are collinear

1. True
2. False

Q. A relation is defined from a set of all triangles on a plane to itself by the rule "is congruent to". The relation is

1. reflexive but not transitive
2. anti symmetric
3. equivalence
4. not symmetric

Q. The function $f:Z\to Z$, $f\left(x\right)=\{\begin{array}{cc}0& \text{if}x\text{is odd}\\ \frac{x}{2}& \text{if}x\text{is even}\end{array}$ then $f$ is

1. surjection but not injection
2. injection but not surjection
3. bijection
4. neither injection nor surjection

Q. Find the position vector of the mid point of the vector joining the points $P(2,3,4)$ and $Q(4,1,2)$.

Q. Let $a,b$ and $c$ are unit vectors such that $a+b+c=0$. Which of the following is correct

1. $a\times b=b\times c=c\times a=0$
2. $a\times b=b\times c=c\times a$ not equal to 0
3. $a\times b=b\times c=a\times c$not equal to 0
4. None of these

Q. If $\overrightarrow{A}=(\overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c})$
$\overrightarrow{B}=(-2\overrightarrow{a}+3\overrightarrow{b}+2\overrightarrow{c})$
$\overrightarrow{C}=(-8\overrightarrow{a}+13\overrightarrow{b})$
Then $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are collinear

1. True
2. False

Q. If $\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{BO}+\overrightarrow{OC}$, then

1. A, B and C are collinear
2. A, B and C are not collinear
3. Nothing can be said about Collinearity from this information
4. $\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}$ form a triangle