Advertisements
Advertisements
Question
Show that the points whose position vectors are \[- 2 \hat{i} + 3 \hat{j} , \hat{i} + 2 \hat{j} + 3 \hat{k} \text{ and } 7 \text{ i} - \text{ k} \] are collinear.
Advertisements
Solution
Let the given points be P, Q and R and let their position vectors be \[\overrightarrow{a} , \overrightarrow{b} \text { and } \overrightarrow{c} , \text{ respectively } . \]
\[\overrightarrow{a} = - 2 \hat{i} + 3 \hat{j} \]
\[ \overrightarrow{b} = \hat{i} + 2 \hat{j} + 3 \hat{k} \]
\[ \overrightarrow{c} = 7 \hat{i} + 9 \hat{k} \]
Vector equation of line passing through P and Q is
\[\overrightarrow{r} = \overrightarrow{a} + \lambda\left( \overrightarrow{b} - \overrightarrow{a} \right)\]
\[ \Rightarrow \overrightarrow{r} = \left( - 2 \hat{i} + 3 \hat{j}
\right) + \lambda\left\{ \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) - \left( - 2 \hat{i} + 3 \hat{j} \right) \right\}\]
\[ \Rightarrow \vec{r} = \left( - 2 \hat{i} + 3 \hat{j} \right) + \lambda\left( 3 \hat{i} - \hat{j} + 3 \hat{k} \right) . . . (1)\]
If points P, Q and R are collinear, then point R must satisfy (1).
\[\text{ Replacing } \overrightarrow{r} \text{ by } \overrightarrow{c} = 7 \hat{i} + 9 \hat{k} \text{ in } (1), \text { we get } \]
\[7 \hat{i} + 9 \hat{k} = \left( - 2 \hat{i} + 3 \hat{j} \right) + \lambda\left( 3 \hat{i} - \hat{j} + 3 \hat{k} \right)\]
Comparing the coefficients of \[\hat{i} , \hat{j} \text{ and } \hat{k}\]
we get
\[7 = - 2 + 3\lambda, 0 = 3 - \lambda, 9 = 3\lambda\]
∴ \[\lambda\] = 3
These three equations are consistent, i.e. they give the same value of \[\lambda\] Hence, the given three points are collinear.
Disclaimer: The question given in the book has a minor error. The third vectors should be \[7 \hat{i }+ 9 \hat{k} \] The solution here is created accordingly.
APPEARS IN
RELATED QUESTIONS
If a line drawn from the point A( 1, 2, 1) is perpendicular to the line joining P(1, 4, 6) and Q(5, 4, 4) then find the co-ordinates of the foot of the perpendicular.
Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.
If the Cartesian equations of a line are ` (3-x)/5=(y+4)/7=(2z-6)/4` , write the vector equation for the line.
Find the value of p, so that the lines `l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5 ` are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, – 4) and parallel to line l1.
Find the vector and cartesian equations of the line passing through the point (2, 1, 3) and perpendicular to the lines
`(x-1)/1=(y-2)/2=(z-3)/3 and x/(-3)=y/2=z/5`
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector `3hati+2hatj-2hatk`.
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector `2hati -hatj+4hatk` and is in the direction `hati + 2hatj - hatk`.
The Cartesian equation of a line is `(x-5)/3 = (y+4)/7 = (z-6)/2` Write its vector form.
Show that the lines `(x-5)/7 = (y + 2)/(-5) = z/1` and `x/1 = y/2 = z/3` are perpendicular to each other.
Find the vector equation of the lines which passes through the point with position vector `4hati - hatj +2hatk` and is in the direction of `-2hati + hatj + hatk`
ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, \[4 \hat{ i} + 5 \hat{j} -10 \hat{k} , 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \text{ and } - \hat{i} + 2 \hat{j} + \hat{k} .\] Find the vector equation of the line BD. Also, reduce it to cartesian form.
Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector \[\hat{i} - 2 \hat{j} + 3 \hat{k} .\] Reduce the corresponding equation in cartesian from.
Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are \[\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 2}\] Also, reduce the equation obtained in vector form.
Show that the three lines with direction cosines \[\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}\] are mutually perpendicular.
Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and, (1, 2, 5).
Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by \[\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} .\]
Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text { and }\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] are perpendicular to each other.
Find the angle between the following pair of line:
\[\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} \text{ and } \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}\]
Find the angle between the following pair of line:
\[\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1} \text{ and } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}\]
Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]
Find the direction cosines of the line
\[\frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6}\] Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.
Determine whether the following pair of lines intersect or not:
\[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
Find the foot of perpendicular from the point (2, 3, 4) to the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, find the perpendicular distance from the given point to the line.
Find the foot of the perpendicular from (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \lambda - 1 \right) \hat{i} + \left( \lambda + 1 \right) \hat{j} - \left( 1 + \lambda \right) \hat{k} \text{ and } \overrightarrow{r} = \left( 1 - \mu \right) \hat{i} + \left( 2\mu - 1 \right) \hat{j} + \left( \mu + 2 \right) \hat{k} \]
Find the shortest distance between the following pairs of parallel lines whose equations are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} + \hat{j} - \hat{k} \right) + \mu\left( 4 \hat{i} - 2 \hat{j} + 2 \hat{k} \right)\]
Write the vector equations of the following lines and hence determine the distance between them \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} \text{ and } \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}\]
Find the shortest distance between the lines \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} - \hat{j} - \hat{k} + \mu\left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right)\]
Write the vector equation of a line given by \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]
The direction ratios of the line perpendicular to the lines \[\frac{x - 7}{2} = \frac{y + 17}{- 3} = \frac{z - 6}{1} \text{ and }, \frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 4}{- 2}\] are proportional to
If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =
If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
Show that the lines \[\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} \text { and } \frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\] are coplanar.
Choose correct alternatives:
The difference between the slopes of the lines represented by 3x2 - 4xy + y2 = 0 is 2
If 2x + y = 0 is one of the line represented by 3x2 + kxy + 2y2 = 0 then k = ______
Equation of a line passing through (1, 1, 1) and parallel to z-axis is ______.
For the point \[(5,2,-4)\], what is its position vector \[\vec{a}\]?
Which Cartesian equation is obtained for the line through \[(5,2,-4)\] parallel to \[3\hat{i}+2\hat{j}-8\hat{k}\]?
