Continuous practice using ISC Class 12 Maths OP Malhotra Solutions Chapter 23 Three Dimensional Geometry Ex 23(c) can lead to a stronger grasp of mathematical concepts.
S Chand Class 12 ICSE Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(c)
Question 1.
Find the vector equation of a line which passes through the vector 2 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) and is parallel to the vector 2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\). Find the Cartesian form also.
Answer:
We know that the vector equation of line passing through the point whose position vector is \(\vec{a}\) and || to vector \(\vec{b}\) is \(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\), where λ be the some scalar.
Thus required vector eqn. of line through
\(\vec{a}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) and parallel to \(\vec{b}\) = 2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\) is given by
\(\vec{r}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) + 5 \(\hat{k}\) + λ(2 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\))
Let \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point on line (1).
∴ eqn. (1) becomes; x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)
= (3 + 2λ) \(\hat{i}\) + (4 + 2λ) \(\hat{j}\) + (5 – 3λ) \(\hat{k}\)
Comparing the coefficients of \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) on both sides, we have
x = 3 + 2λ
⇒ \(\frac{x-3}{2}\) = λ
y = 4 + 2λ
⇒ \(\frac{y-4}{2}\) = λ
z = 5 – 5λ
⇒ \(\frac{z-5}{-3}\) = λ
Thus \(\frac{x-3}{2}\) = \(\frac{y-4}{2}\) = \(\frac{z-5}{-3}\) be the required eqn. of line in cartesian form.
Question 2.
Find the vector equation of a line which is parallel to the vector 2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\) and which passes through the point (5, -2, 4). Reduce it to the Cartesian form.
Answer:
We know that vector equation of line passes through the point with position vector \(\vec{a}\) and parallel to vector \(\vec{b}\) is given by
\(\vec{r}\) = \(\vec{a}\) + λ \(\vec{b}\)
Here \(\vec{a}\) = 5 \(\hat{i}\) – 2 \(\hat{j}\) + 4 \(\hat{k}\)
and \(\vec{b}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)
Thus the required vector eqn. of line be \(\vec{r}\) = 5 \(\hat{i}\) – 2 \(\hat{j}\) + 4 \(\hat{k}\) + λ(2\(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\))
Let \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the position vector of any point on this line
∴ From (1) we get x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)
= 5 \(\hat{i}\) – 2 \(\hat{j}\) + 4 \(\hat{k}\) + λ(2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\))
⇒ x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)
= (5 + 2 λ) \(\hat{i}\) + (-2 – λ) \(\hat{j}\) + (4 + 3λ) \(\hat{k}\)
On equating the coefficients of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) on both sides
we have x = 5 + 2λ, y = -2 – λ and z = 4 + 3λ
⇒ \(\frac{x-5}{2}\) = λ; \(\frac{y+2}{-1}\) = λ;
\(\frac{z-4}{3}\) = λ
i.e. \(\frac{x-5}{2}\) = \(\frac{y+2}{-1}\) = \(\frac{z-4}{3}\)
[on eliminating λ] be the required eqn. of line in cartesian form.
Question 3.
Find the vector and Cartesian equations of the line that passes through the points (3, -2, -5), (3, -2, 6).
Answer:
Let \(\vec{a}\) and \(\vec{b}\) be the position vectors of the given points A(3, -2, -5) and B(3, -2, 6)
Thus required vector eqn. of line through the points A and B with P.V. \(\vec{a}\) and \(\vec{b}\) is given by
\(\vec{r}\) = \(\vec{a}\) + λ(\(\vec{b}\) – \(\vec{a}\))
∴ \(\vec{r}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) – 5 \(\hat{k}\) +
λ[(3 \(\hat{i}\) – 2 \(\hat{j}\) + 6 \(\hat{k}\)) – (3 \(\hat{i}\) – 2 \(\hat{j}\) – 5 \(\hat{k}\))]
⇒ \(\vec{r}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) – 5 \(\hat{k}\) + λ(0 \(\hat{i}\) + 0 \(\hat{j}\) + 11 \(\hat{k}\))
⇒ \(\vec{r}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) + (11λ – 5) \(\hat{k}\)
The cartesian eqn. of line be
\(\frac{x-3}{3-3}\) = \(\frac{y(-2)}{-2+2}\) = \(\frac{z-(-5)}{6-(-5)}\)
i.e., \(\frac{x-3}{0}\) = \(\frac{y+2}{0}\) = \(\frac{z+5}{11}\)
Question 4.
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2 \(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\) and is in the direction \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\)
Answer:
We know that vector eqn. of line passing through the point whose P.V. \(\vec{a}\) and || to vector \(\vec{b}\) is
\(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)
Thus required vector eqn. of line through the point with P.V. 2 \(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\) and in the direction of \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\) is
\(\vec{r}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\) + λ(\(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\))
Let \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point on line (1)
∴ eqn. (1) becomes; x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) = (2 + λ) \(\hat{i}\) + (-1 + 2λ) \(\hat{j}\) + (4-λ) \(\hat{k}\)
On comparing the coefficients of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) on both sides, we have
x = 2 + λ
⇒ \(\frac{x-2}{1}\) = λ;
z = 4 – λ
⇒ \(\frac{z-4}{-1}\) = λ
y = -1 + 2λ
⇒ \(\frac{y+1}{2}\) = λ
Thus, \(\frac{x-2}{1}\) = \(\frac{y+1}{2}\) = \(\frac{z-4}{-1}\) be the required equal line in cartesian form.
Question 5.
The Cartesian equations of a line are 6 x – 2 = 3 y + 1 = 2 z – 2. Find the vector equation of this line.
Answer:
Eqn. of given line be,
6 x – 2 = 3 y + 1 = 2 z – 2
⇒ 6(x – \(\frac{1}{3}\)) = 3(y + \(\frac{1}{3}\)) = 2(z – 1)
⇒ \(\frac{x-1 / 3}{1 / 6}\) = \(\frac{y+1 / 3}{1 / 3}\) = \(\frac{z-1}{1 / 2}\)
⇒ \(\frac{x-1 / 3}{1}\) = \(\frac{y+1 / 3}{2}\) = \(\frac{z-1}{3}\)
∴ D’ratios of given line are proportional to < 1, 2, 3 >.
Since the required line is || to given line
∴ required line || to the vector \(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)
Hence the vector eqn. of line which passing threough the point (\(\frac{1}{3}\), \(\frac{-1}{3}\), 1) and || to vector \(\vec{b}\) is
\(\vec{r}\) = \(\frac{1}{3}\) \(\hat{i}\) – \(\frac{1}{3}\) \(\hat{j}\) + \(\hat{k}\) + λ(\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\))
Question 6.
Find the vector equation of a line passing through the point with position vector \(\hat{i}\) – 2 \(\hat{j}\) – 3 \(\hat{k}\) and parallel to the line joining the points with position vectors \(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\) and 2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\). Also find the Cartesian form of the equation.
Answer:
Let A, B, C be the given points whose position vectors are
\(\hat{i}\) – 2 \(\hat{j}\) – 3 \(\hat{k}\) ; \(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\); 2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)
we want to find the eqn. of line passing through A and parallel to \(\overrightarrow{\mathrm{BC}}\)
Now \(\overrightarrow{\mathrm{BC}}\) = P.V. of C – P. V. of B = (2 \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)) – (\(\hat{i}\) – \(\hat{j}\) + 4 \(\hat{k}\)) = \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\)
also vector eqn. of line passes through point A with P.V. \(\vec{a}\) and || to \(\vec{b}\) is given by \(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)
Here \(\vec{a}\) = \(\hat{i}\) – 2 \(\hat{j}\) – 3 \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\)
i.e. \(\vec{r}\) = \(\hat{i}\) – 2 \(\hat{j}\) – 3 \(\hat{k}\) + λ(\(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\)) be the required vector equation.
putting \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) in this line, we get
x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) = (1 + λ) \(\hat{i}\) + (-2 + 2λ) \(\hat{j}\) + (-3 – 2) \(\hat{k}\)
equating the coefficients of \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) on both sides, we get
∴x = 1 + λ ; y = -2 + 2 λ ; z = -3 – 2 λ
⇒ \(\frac{x-1}{1}\) = λ ; \(\frac{y+2}{2}\) = λ; \(\frac{z+3}{-2}\) = λ
On eliminating λ we get ; \(\frac{x-1}{1}\) = \(\frac{y+2}{2}\) = \(\frac{z+3}{-2}\) be the required cartesian equation of line.
Question 7.
Show that the points, whose position vectors are given by
-4 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\), \(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\) and (-9 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\)) are collinear.
Answer:
Let A, B, C be the points with \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be then refrective position vectors.
Then \(\vec{a}\) = -4 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\); \(\vec{b}\) = \(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\) and \(\vec{c}\) = -9 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\)
∴ vector eqn. of line passing through the points A(\(\vec{a}\)) and B(\(\vec{b}\)) is given by
\(\vec{r}\) = \(\vec{a}\) + λ(\(\vec{b}\) – \(\vec{a}\))
⇒ \(\vec{r}\) = (-4 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) + λ[(\(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\)) – (-4 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\))]
⇒ \(\vec{r}\) = (-4 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) + λ(5 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
Now the given points A, B and C are collinear if point C lies on eqn. (1).
i.e. if point C lies on eqn. (1).
i.e., if P.V. of \(\vec{c}\) i.e. -9 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\) satisfies eqn. (1).
i.e., if -9 \(\hat{i}\) + \(\hat{j}\) – 4 \(\hat{k}\) = (-4 \(\hat{i}\) + 2 \(\hat{j}\) – 3 \(\hat{k}\)) + λ(5 \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
i.e., if -9 = -4 + 5λ, 1 = 2 + λ, -4 = -3 + λ i.e., if λ = -1, λ = -1, λ = -1
Hence, λ = 1 satisfies all equation. Thus the given points are collinears.
Question 8.
The points A(4, 5, 10), B(2, 3, 4) and C(1, 2, -1) are three vertices of a parallelogram A B C D. Find the vector and Cartesian equations for the sides A B and B C and find the coordinates of D.
Answer:
The vector eqn, of line through the point A(\(\vec{a}\)) and B(\(\vec{b}\)) is given by
\(\vec{r}\) = \(\vec{a}\) + λ(\(\vec{b}\) – \(\vec{a}\))
where \(\vec{a}\) = 4 \(\hat{i}\) + 5 \(\hat{j}\) + 10 \(\hat{k}\), \(\vec{b}\)
= 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) and \(\vec{c}\) = \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\) Hence, vector eqn. of side A B is given by
\(\vec{r}\) = 4 \(\hat{i}\) + 5 \(\hat{j}\) + 10 \(\hat{k}\) + λ[(2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) – (4 \(\hat{i}\) + 5 \(\hat{j}\) + 10 \(\hat{k}\))]
i.e. \(\vec{r}\) = 4 \(\hat{i}\) + 5 \(\hat{j}\) + 10 \(\hat{k}\) + λ(-2 \(\hat{i}\) – 2 \(\hat{j}\) – 6 \(\hat{k}\))
and cartesian equ. of AB be given by
\(\frac{x-4}{2-4}\) = \(\frac{y-5}{3-5}\) = \(\frac{z-10}{4-10}\)
i.e., \(\frac{x-4}{-2}\) = \(\frac{y-5}{-2}\) = \(\frac{z-10}{-6}\)
i.e., \(\frac{x-4}{1}\) = \(\frac{y-5}{1}\) = \(\frac{z-10}{3}\)
Therefore, vector eqn. of line BC be given by
\(\vec{r}\) = \(\vec{b}\) + A(\(\vec{c}\) – \(\vec{b}\))
i.e., \(\vec{r}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) + λ(\(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\)) – (2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\))]
i.e., \(\vec{r}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) + λ(\(-\hat{i}\) – \(\hat{j}\) – 5 \(\hat{k}\))
i.e., \(\vec{r}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) + λ(\(\hat{i}\) + \(\hat{j}\) + 5 \(\hat{k}\));
where λ = -λ
Hence, required cartesian equation of line C D is given by
\(\frac{x-2}{1-2}\) = \(\frac{y-3}{2-3}\) = \(\frac{z-4}{-1-4}\)
i.e., \(\frac{x-2}{-1}\) = \(\frac{y-3}{-1}\) = \(\frac{z-4}{-5}\)
i.e., \(\frac{x-2}{1}\) = \(\frac{y-3}{1}\) = \(\frac{z-4}{5}\)
LetD(α, β, γ) be the coordinates of vertex D of || gm ABCD. Since diagonals AC and BD bisect each other.
∴ Mid point of AC = mid point of BD
i.e., (\(\frac{4+1}{2}\), \(\frac{5+2}{2}\), \(\frac{10-1}{2}\))
= (\(\frac{2+\alpha}{2}\), \(\frac{3+\beta}{2}\), \(\frac{4+\gamma}{2}\))
i.e., \(\frac{5}{2}\) = \(\frac{2+\alpha}{2}\)
⇒ α = 3 ; \(\frac{7}{2}\) = \(\frac{3+\beta}{2}\)
⇒ β = 4 ; \(\frac{9}{2}\) = \(\frac{4+\gamma}{2}\)
⇒ γ = 5
Thus, required coordinates of D are (3, 4, 5).
Question 9.
Find the Cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line \(\frac{-x-2}{1}\) = \(\frac{y+3}{7}\) = \(\frac{2 z-6}{3}\)
Answer:
eqn. of given line be \(\frac{-x-2}{1}\) = \(\frac{y+3}{7}\) = \(\frac{2 z-6}{3}\)
i.e. \(\frac{x+2}{-1}\) = \(\frac{y+3}{7}\) = \(\frac{z-3}{\frac{3}{2}}\)
∴ D’ratios of given line (1) proportional to < -1, 7, > \(\frac{3}{2}\)
i.e. < -2, 14, 3 >
Thus Direction ratios of required line i.e. line || to line (1) are proportional to < -2, 14, 3 >
Hence the catesian eqn. of line passing through (1, 2, 3) having direction ratios < -2, 14, 3 > is given by \(\frac{x-1}{-2}\) = \(\frac{y-2}{14}\) = \(\frac{z-3}{3}\) = t (say)
⇒ x = -2 t + 1 ; y = 14 t + 2 ; z = 3 t + 3
Let \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be P.V. of any point on line
Then \(\vec{r}\) = (-2 t + 1) \(\hat{i}\) + (14 t + 2) \(\hat{j}\) + (3 t + 3) \(\hat{k}\)
⇒ \(\vec{r}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\) + t(-2 \(\hat{i}\) + 14 \(\hat{j}\) + 3 \(\hat{k}\)) be the required vector eqn. of line.
Question 10.
Find the direction cosines and vector equation of the line whose Cartesian form is \(\frac{x-2}{2}\) = \(\frac{2 y-5}{-3}\), z = -1.
Answer:
Given eqn. of line in cartesian form be
\(\frac{x-2}{2}\) = \(\frac{2 y-5}{-3}\) = \(\frac{z+1}{0}\)
i.e., \(\frac{x-2}{2}\) = \(\frac{y-5 / 2}{-3 / 2}\) = \(\frac{z+1}{0}\)
∴ eqn. of given lines be passes through the point (2, \(\frac{5}{2}\), -1)
and having direction ratios < 2, \(\frac{-3}{2}\), 0 >
i.e., < 4, -3, 0 >
Thus, vector eqn. of line passes through the point (2, \(\frac{5}{2}\), -1) whose P.V. be \(\vec{a}\) = 2 \(\hat{i}\) + \(\frac{5}{2}\) \(\hat{j}\) – \(\hat{k}\) and || to vector \(\vec{b}\) = 4 \(\hat{i}\) -3 \(\hat{j}\) + 0 \(\hat{k}\) is given by
\(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)
i.e., \(\vec{r}\) = 2 \(\hat{i}\) + \(\frac{5}{2}\) \(\hat{j}\) – \(\hat{k}\) + λ(4 \(\hat{i}\) – 3 \(\hat{j}\) + 0 \(\hat{k}\))
∴ direction cosines of line be < \(\frac{4}{\sqrt{4^2+(-3)^2+0^2}}\), \(\frac{-3}{\sqrt{4^2+(-3)^2+0}}\), 0 >
i.e., < \(\frac{4}{5}\), \(\frac{-3}{5}\), 0 >