These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise
Question 1.
Show that the line joining the origin to the point (2,1,1) is perpendicular to the line de-termined by the points (3, 5, – 1), (4, 3, -1).
Solution:
The direction ratios of the line joining (0, 0, 0) and (2, 1, 1) are 2, 1,1 The direction ratios Of the line joining (3, 5, -1) and (4, 3, -1) are 1, -2, 0
∴ a1a2 + b1b1 + C1C2
= 2(1) + 1(-2) + 1 (0) = 0
The lines are perpendicular.
Question 2.
If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m1n2 – m2 n1, n1 l2 – n2 l1, l1 m2 – l2 m1
Solution:.
Let \(\hat{a}\) and \(\hat{b}\) be unit vectors in the direction of the lines
∴ \(\hat{a}=l_{1} \hat{i}+m_{1} \hat{j}+n_{1} \hat{k}\)
\(\hat{b}=l_{2} \hat{i}+m_{2} \hat{j}+n_{2} \hat{k}\)
Since a and b are mutually perpendicular, then \(\hat{a}\) x \(\hat{b}\) is a unit vector perpendicular to \(\hat{a}\) and \(\hat{b}\)
∴ The direction cosines are m1n2 – m2 n1, n1 l2 – n2 l1, l1 m2 – l2 m1
Question 3.
Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.
Solution:
a1 = a, b1 = b, c1 = c
a2 = b – c, b2 = c – a, c2 = a – b
Let θ be the angle between the lines
Question 4.
Find the equation of a line parallel to x- axis and passing the origin.
Solution:
The direction ratios of any line parallel to x-axis is 1, 0, 0. Hence the equation of the line passing through the origin having direction ratios 1,0,0 is
\(\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}\)
i.e., \(\frac{x}{1}=\frac{y}{0}=\frac{z}{0}\)
Question 5.
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (- 4, 3, -6) and (2, 9, 2) respectively, then find the angle between the line AB and CD.
Solution:
The direction ratios of AB
= 4 – 1, 5 – 2, 7 – 3 = 3, 3, 4
The direction ratios of CD = 6, 6, 8
Let θ be the angle between AB and CD
Question 6.
If the lines \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are perpendicular, find the value of k.
Solution:
The equation of the lines are
\(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and
\(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\)
The direction ratios of 1st line = – 3, 2k, 2
The direction ratios of IInd line = 3k, 1, – 5
Since the two lines are perpendicular,
a1a2 + b1b2 + c1c2 = 0
– 3(3k) + 2k(1) + 2(- 5) = 0
– 9k + 2k – 10 = 0
– 7k = 10
k = \(\frac { -10 }{ 7 }\)
Question 7.
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \(\vec{r} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})+9\) = 0.
Solution:
Let \(\hat{a}\) be the position vector of the point (1, 2, 3)
a = \(\hat{i}+2 \hat{j}+3 \hat{k}\)
The direction ratios of the normal to the plane are 1, 2, – 5.
Let b = \(\hat{i}+2 \hat{j}-5 \hat{k}\)
Let \(\hat{r}\) be the position vector of any point on the line.
The vector equation is \(\vec{r}=\vec{a}+\lambda \vec{b}\)
i.e., \(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+2 \hat{j}-5 \hat{k})\)
Question 8.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 2
Solution:
The given plane is \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 2
i.e., x + y + z – 2
The equation of any plane parallel to the given plane is x +y + z = k … (1)
Since the required plane passes through the point (a, b, c), we get a + b + c = k
Thus (1) gives the required plane as x + y + z = a + b + c
Question 9.
Find the shortest distance between the lines \(\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})\) and \(\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})\)
Solution:
Question 10.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.
Solution:
The equation of the line through the points (5,1, 6) and (3,4, 1) is
\(\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}=\lambda\)
i.e., \(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda\)
i.e., x = – 2λ. + 5, y = 3λ + 1, z = – 5λ + 6
Since this line crosses the YZ – plane, x coordinate is zero
i.e., – 2λ + 5 = 0 ∴ λ = \(\frac { 5 }{ 2 }\)
\(y=3\left(\frac{5}{2}\right)+1=\frac{17}{2}\)
\(z=-5\left(\frac{5}{2}\right)+6=\frac{-13}{2}\)
∴ The required point is \(\left(0, \frac{17}{2}, \frac{-13}{2}\right)\)
Question 11.
Find the coordinates of the point where the line through (5, 1,6) and (3,4, 1) crosses the ZX- plane.
Solution:
The equation of the line through (5, 1, 6) and (3, 4, 1) is
\(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda\)
i.e., x = – 2λ + 5, y = 3λ + 1, z = – 5λ + 6
Since this line crosses the ZX – plane,
y coordinate is zero.
Question 12.
Find the coordinates of the point where the line through (3, – 4, -5) and (2, -3, 1) crosses the plane 2x + y + z = 7.
Solution:
The equation of the line through (3, – 4, – 5) and (2, – 3, 1) is
The coordinate of any point P on the line is (- λ + 3, λ – 4, 6λ – 5)
P is a point on the plane 2x + y + z = 7
i.e., 2(- λ + 3) + (λ – 4) + (6λ – 5) = 7
– 2λ + 6 + λ – 4 + 6λ – 5 = 7
5λ – 3 = 7 ∴ λ = 2
∴ P is (- 2+ 3, 2 – 4, 6(2) – 5) = (1, – 2, 7)
∴ The coordinates of the point which crosses the plane is (1, -2, 7)
Question 13.
Find the equation of the plane passing through the point (- 1, 3, 2) and perpendicular to each of the planes x + 2y +3z = 5 and 3x + 3y + z = 0.
Solution:
The plane passes through (- 1, 3, 2)
∴ x1 = – 1, y1 = 3, z1 = 2
The direction ratios of the normal to the planes are 1, 2, 3 and 3, 3, 1.
Hence the equation of the plane is
\(\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|\) = 0
\(\left|\begin{array}{ccc} x+1 & y-3 & z-2 \\ 1 & 2 & 3 \\ 3 & 3 & 1 \end{array}\right|\) = 0
– 7(x + 1) + 8(y – 3) – 3(z – 2) = 0
i.e., – 7x – 7 + 8y – 24 – 3z + 6 = 0
– 7x + 8y – 3z – 25 = 0
7x – 8y + 3z + 250 = 0
Another method:
The equation of a plane passing through the point(- 1, 3, 2) is
a(x + 1) + b(y – 3) + c(z – 2) = 0 … (1)
where a, b and c are the direction ratios of the normal to the plane.
Since (1) is perpendicular to the planes
x + 2y + 3z = 5 and 3x + 3y + z = 0, we get
a + 2b + 3c = 0 … (2) and
3a + 3b + c = 0 … (3)
∴ By the rule of cross multiplication, we get
\(\frac{a}{\left|\begin{array}{ll} 2 & 3 \\ 3 & 1 \end{array}\right|}=\frac{0}{\left|\begin{array}{ll} 3 & 1 \\ 1 & 3 \end{array}\right|}=\frac{c}{\left|\begin{array}{ll} 1 & 2 \\ 3 & 3 \end{array}\right|}\)
i.e., \(\frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}\)
Since a, b, c are the direction ratios of the normal to the plane, we can take a = – 7, b = 8, c = – 3.
∴ (1) → 7(x+ 1) + 8(y – 3) – 3(z – 2) = 0
i.e., – 7x + 8y – 3z – 25 = 0
or 7x – 8y + 3z + 25 = 0
Question 14.
If the points (1, 1, p) and (-3, 0, 1) are equidistant from the plane \(\vec{r} \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})+13\) = 0, then find the value of p.
Solution:
The equation of the plane is,
Question 15.
Find the equation of the plane passing through the line of intersection of the planes \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 1 and \(\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4\) = 0 and parallel to x- axis.
Solution:
Equation of the given planes are
x + y + z – 1 = 0 and 2x + 3y – z + 4 = 0
The equation of the plane passing through the line of intersection of the planes is
(x + y + z – 1) + (2x + 3y – z + 4) = 0
(1 + 2λ)x + (1 + 3λ)y + (1 – λ )z + (4 – 1) = 0 … (1)
The direction ratios of the normal to the plane are 1 + 2λ , 1 + 3λ , 1 – λ
The direction ratios of x-axis are 1, 0, 0
Since the required plane is parallel to the x-axis, the normal to the plane is perpendicular to x-axis.
Question 16.
If O is the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.
Solution:
\(\overrightarrow{\mathrm{OP}}=\hat{i}+2 \hat{j}-3 \hat{k}\)
Direction ratios of OP are 1, 2, -3
OP is perpendicular to the plane
∴ Direction ratios of the normal to the plane
are a = 1, b = 2, c = -3
P( 1, 2, – 3) is a point on the plane.
Equation of the plane is
a(x – x1) + b(y – y1) + c(z – z1) = 0
i.e, 1(x – 1) + 2(y – 2) – 3(z + 3) = 0
x – 1 + 2y – 4 – 3z – 9 = 0
x + 2y – 3z – 14 = 0
The equation of the plane is x + 2y – 3z – 14 = 0
Question 17.
Find the equation of the plane which contains the line of intersection of the planes \(\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4\) = 0, \(\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5\) = 0 and which is perpendicular to the plane \(\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8\) = 0.
Solution:
The equation of the given planes are x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
The equation of the plane containing the line of inÍersection of the above plane is
(x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0
(1 + 2λ)x + (2 + λ)y (3 – λ)z + (5λ – 4) = 0 … (1)
Since (1) is perpendicular to the plane
5x + 3y – 6z + 8 = 0, we get
5(1 + 2λ) + 3(2 + λ) – 6(3 – λ) = 0
5 + 10λ + 6 + 3λ – 18 + 6λ = 0
19λ – 7 = 0
∴ λ = \(\frac { 7 }{ 19 }\)
Question 18.
Find the distance of the point (- 1, – 5, – 10) from the point of intersection of the line \(\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})\) = 5
Solution:
The cartesian equation of the line \(\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})\) is
\(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}\) = λ … (1)
The cartesian equation of the plane
\(\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})\) is x – y + z = 5 … (2)
The coordinate of any point on this line is P(2 + 3λ, – 1 + 4λ, 2 + 2λ).
Let the line intersects the plane at P, then P is a point on the plane.
∴ (2) → (2 + 3λ) – (-1 + 4λ) + (2 + 2λ) = 5
2 + 3λ + 1 – 4λ + 2 + 2λ = 5
∴ λ = 0
∴ P is (2, – 1,2) and the given point is Q(- 1, – 5, – 10)
∴ PQ = \(\sqrt{(-1-2)^{2}+(-5–1)^{2}+(-10-2)^{2}}\)
= \(\sqrt{(-3)^{2}+(-4)^{2}+(-12)^{2}}\)
= \(\sqrt{9+16+144}=\sqrt{169}\)
= 13
Question 19.
Find the vector equation of the line passing through (1,2,3) and parallel to the planes \(\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=5 \text { and } \vec{r} \cdot(3 \hat{i}+\hat{j}+\hat{k})=6\)
Solution:
The planes are \(\hat{r}\). \(\hat{n}\)2 = d1
and \(\hat{r}\). \(\hat{n}\)2 = d2
where \(\hat{n}\)1 = \(\hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{n}_{2}=3 \hat{i}+\hat{j}+\hat{k}\)
Since the required line is parallel to the planes, it is perpendicular to both \(\hat{n}\)1 and \(\hat{n}\)2.
\(\hat{n}\)1 x \(\hat{n}\)2 is perpendicular to both \(\hat{n}\)1 and \(\hat{n}\)2
\(\vec{n}_{1} \times \vec{n}_{2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 2 \\
3 & 1 & 1
\end{array}\right|=-3 \hat{i}+5 \hat{j}+4 \hat{k}\)
The required line passes through the point (1, 2, 3) and is parallel to \(\hat{n}\)1 x \(\hat{n}\)2.
∴ The equation of the line is
\(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-3 \hat{i}+5 \hat{j}+4 \hat{k})\)
Question 20.
Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:
\(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\)
\(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\)
Solution:
Let \(\vec{b}_{1}\) and \(\vec{b}_{2}\) be the direction of the lines where \(\vec{b}_{1}=3 \hat{i}-16 \hat{j}+7 \hat{k}\) and \(\vec{b}_{2}=3 \hat{i}+8 \hat{j}-5 \hat{k}\).
The required line is perpendicular to both \(\vec{b}_{1}\) and \(\vec{b}_{2}\). Hence it is parallel to \(\vec{b}_{1}\) x \(\vec{b}_{2}\)
\(\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -16 & 7 \\
3 & 8 & -5
\end{array}\right|=24 \hat{i}+36 \hat{j}+72 \hat{k}\)
The direction ratios of \(\vec{b}_{1}\) x \(\vec{b}_{2}\) = 24, 36, 72
= 2, 3, 6
The required line passes through the point (1, 2, – 4 ) and has direction ratios 2, 3, 6.
Hence the equation is
\(\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\)
Question 21.
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then \(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}\).
Solution:
The equation of a plane in the intercept from is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1.
Question 22.
Distance between the two planes:
2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
a. 2 units
b. 4 units
c. 8 units
d. \(\frac{2}{\sqrt{29}}\) units
Solution:
d. \(\frac{2}{\sqrt{29}}\) units
Hence the planes are parallel.
The distance between the parallel planes is the difference between their distance from the origin.
The distance of the plane 2x + 3y + 4z = 4
Question 23.
The planes 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
a. perpendicular
b. parallel
c. intersect y-axis
d. passes through (0, 0, \(\frac { 5 }{ 4 }\))
Solution:
b. parallel
