Equation of a Line Passing through Two Given Points

Q. Let $L$ be the line of intersection of planes $\overrightarrow{r}\cdot (\hat{i}-\hat{j}+2\hat{k})=2$ and $\overrightarrow{r}\cdot (2\hat{i}+\hat{j}-\hat{k})=2$. If $P(\alpha ,\beta ,\gamma )$ is the foot of perpendicular on $L$ from the point $(1,2,0)$, then the value of $35(\alpha +\beta +\gamma )$ is equal to

1. $143$
2. $101$
3. $134$
4. $119$

Q. A line $L$ passing through origin is perpendicular to the lines
$L_1:\overrightarrow{r}=(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k}$
$L_2:\overrightarrow{r}=(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k}$
If the co-ordinates of the point in the first octant on $L_2$ at the distance of $\sqrt{17}$ from the point of intersection of $L$ and $L_1$ are $(a,b,c),$ then $18(a+b+c)$ is equal to

Q. Let $Q$ be the foot of the perpendicular from the point $P(7,-2,13)$ on the plane containing the lines $\frac{x+1}{6}=\frac{y-1}{7}=\frac{z-3}{8}$ and $\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}.$ Then $(PQ{)}^2,$ is equal to

Q. If the shortest distance between the straight lines $3(x-1)=6(y-2)=2(z-1)$ and $4(x-2)=2(y-\lambda )=(z-3),\lambda \in R$ is $\frac{1}{\sqrt{38}},$ then the integral value of $\lambda$ is equal to:

1. $2$
2. $5$
3. $3$
4. $-1$

Q. The line passing through the points (5, 1, a) and (3, b, 1) crosses the yz-plane at the point $(0,\frac{17}{2},\frac{-13}{2}),$ Then

1. a = 2, b = 8
2. a = 4, b = 6
3. a = 6, b = 4
4. a = 8, b = 2

Q. A line L passing through the point $P(1,4,3)$ is perpendicular to both the lines
$\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-2}{4}\text{and}\frac{x+2}{3}=\frac{y-4}{2}=\frac{z+1}{-2}$.
If the position vector of the point Q on L is $(a_1,a_2,a_3)$ such that $P{Q}^2=357,\text{then}(a_1+a_2+a_3)$ can be

1. 15
2. 16
3. 2
4. 1

Q. If the acute angle between the line $\overrightarrow{r}=\hat{i}+2\hat{j}+\lambda (4\hat{i}-3\hat{k})$ and $xy-$plane is $\alpha$ and the acute angle between the planes $x+2y=0$ and $2x+y=0$ is $\beta ,$ then $({\cos }^2\alpha +{\sin }^2\beta )$ equals

Q. If a point $R(4,y,z)$ lies on the line segment joining the points $P(2,-3,4)$ and $Q(8,0,10)$, then the distance of $R$ from the origin is :

1. $\sqrt{53}$
2. $2\sqrt{14}$
3. $2\sqrt{21}$
4. $6$

Q. The plane 4x + 7y + 4z + 81 = 0 is rotated through a right angle about its line of intersection with the plane 5x + 3y + 10z = 25. The equation of the plane in its new position is x – 4y +6z = k, where k is

1. 73
2. 37
3. -89
4. 106

Q.

Find the distance of the point (- 1, - 5, - 10) from the point of intersection of the line $r=2\hat{i}-\hat{j}+2\hat{k}+\lambda (3\hat{i}+4\hat{j}+2\hat{k})$ and the plane $r.(\hat{i}-\hat{j}+\hat{k})=5$

Q.

Find the Cartesian equation of the line which passes through the point (- 2, 4, -5) and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$

Q. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ -plane.

Q. Find the distance of the point (-2, 3, -4) from the line
x+2/3 = 2y+3/4=3z+4/5 measure parallel to the plane
4x+12y-3z+1=0.

Q. The distance between the line $x=2+t,$ $y=1+t,$ $z=\frac{-1}{2}-\frac{t}{2}$ where $t\in R$ and the plane $\overrightarrow{r}\cdot (\hat{i}+2\hat{j}+6\hat{k})=10$ is

1. $\frac{1}{6}$
2. $\frac{1}{\sqrt{41}}$
3. $\frac{1}{7}$
4. $\frac{9}{\sqrt{41}}$

Q. Find the equation of the line through the points (3, 4, -7) and (1, -1, 6)

1. 
2. 
3. 
4. 

Q. Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, −4, −5) and B(2, −3, 1) intersects the plane 2x + y + z = 7. [CBSE 2015]

Q.

Find the coordinates of the point where the line through (3, ­−4, −5) and (2, − 3, 1) crosses the plane 2x + y + z = 7).

Q.

If 0 is the origin and A is (a, b, c), then find the direction cosines of the line OA and the equation of plane through A at right angle OA.

Q.

Find the orthocecentre of the triangle with the vertices (-5, -7), (13, 2), (-5, 6)

Q.

Determine the direction cosines of the normal to plane and the distance from the origin:

5y + 8 = 0

Q.

P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is

Q. The distance of the point $P(3,4,4)$ from the point of intersection of the line joining the points $Q(3,–4,–5)$ and $R(2,–3,1)$ and the plane
$2x+y+z=7$, is equal to

Q. A line L passing through the point P(1, 4, 3) is perpendicular to both the lines $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-2}{4}$ and $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z+1}{-2}$ . If the position vector of the point Q on L is $(a_1,a_2,a_3)$ such that $P{Q}^2=357$, then $(a_1+a_2+a_3)$ can be

1. 15
2. 1
3. 2
4. 16

Q.

Find the equation of the plane which contains line of intersection of the planes $r.(\hat{i}+2\hat{j}+3\hat{k})-4=0,r.(2\hat{i}+\hat{j}-\hat{k})+5=0$ and which is perpendicular to the plane $r.(5\hat{i}+3\hat{j}-6\hat{k})+8=0$ .

Q. Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

Q. Using differentials, find the approximate value of each of the following up to $3$ places of decimal.
(i) $\sqrt{25.3}$
(ii) $\sqrt{49.5}$
(iii) $\sqrt{0.6}$
(iv) $(0.009{)}^{\frac{1}{3}}$
(v) $(0.999{)}^{\frac{1}{10}}$
(vi) $(15{)}^{\frac{1}{4}}$
(vii) $(26{)}^{\frac{1}{3}}$
(viii) $(255{)}^{\frac{1}{4}}$
(ix) $(82{)}^{\frac{1}{4}}$
(x) $(401{)}^{\frac{1}{2}}$
(xi) $(0.0037{)}^{\frac{1}{2}}$

Q. The coordinates of the foot of the perpendicular from the points $(1,-2,1)$ on the plane containing the lines, $\frac{x+1}{6}=\frac{y-1}{7}=\frac{z-3}{8}$ and $\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}$ is:

1. $(2,-4,2)$
2. $(0,0,0)$
3. $(-1,2,-1)$
4. $(1,1,1)$

Q. Let P be the foot of the perpendicular dropped from the origin O on the line of intersection of the planes $x-2y+3z=5$ and $2x+3y+z+4=0$, then

1. Equation of the plane perpendicular to the given line and passing through the point (1, 1, 1) is $11x-5y-7z+1=0$
2. Equation of acture angle bisector of the two planes is $x+5y-2z+9=0$.
3. $\overline{OP},\hat{i}-2\hat{j}+3\hat{k}$ and $2\hat{i}+3\hat{j}+\hat{k}$ must be coplanar
4. The point (1, 2, 3) lies in acute region of the two planes

Q. If the distance of the point $(1,-2,3)$ from the plane $x+2y-3z+10=0$ measured parallel to the line, $\frac{x-1}{3}=\frac{2-y}{m}=\frac{z+3}{1}$ is $\sqrt{\frac{7}{2}},$ then the value of $|m|$ is equal to

Q. Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).