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Access Three Dimensional Geometry NCERT Solutions Exercise 11.2 Chapter-11 Class 12
NCERT Question 1: Show that the three lines with direction cosines
$$\left(\frac{12}{13}, -\frac{3}{13}, -\frac{4}{13}\right);
\left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right);
\left(\frac{3}{13}, -\frac{4}{13}, \frac{12}{13}\right)$$
are mutually perpendicular.
Solution
Two lines with direction cosines $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are perpendicular to each other if
$$l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$$
(i) For the lines with direction cosines
$$\left(\frac{12}{13}, -\frac{3}{13}, -\frac{4}{13}\right) , \left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right)$$
$$l_1l_2 + m_1m_2 + n_1n_2
= \frac{12}{13}\cdot\frac{4}{13} + \left(-\frac{3}{13}\right)\cdot\frac{12}{13} + \left(-\frac{4}{13}\right)\cdot\frac{3}{13}$$
$$l_1l_2 + m_1m_2 + n_1n_2= \frac{48}{169} – \frac{36}{169} – \frac{12}{169} = 0$$
Hence, the lines are perpendicular.
(ii) For the lines with direction cosines
$$\left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right) , \left(\frac{3}{13}, -\frac{4}{13}, \frac{12}{13}\right)$$
$$l_1l_2 + m_1m_2 + n_1n_2
= \frac{4}{13}\cdot\frac{3}{13} + \frac{12}{13}\cdot\left(-\frac{4}{13}\right) + \frac{3}{13}\cdot\frac{12}{13}$$
$$l_1l_2 + m_1m_2 + n_1n_2= \frac{12}{169} – \frac{48}{169} + \frac{36}{169} = 0$$
Hence, the lines are perpendicular.
(iii) For the lines with direction cosines
$$\left(\frac{3}{13}, -\frac{4}{13}, \frac{12}{13}\right) , \left(\frac{12}{13}, -\frac{3}{13}, -\frac{4}{13}\right)$$
$$l_1l_2 + m_1m_2 + n_1n_2
= \frac{3}{13}\cdot\frac{12}{13} + \left(-\frac{4}{13}\right)\cdot\left(-\frac{3}{13}\right) + \frac{12}{13}\cdot\left(-\frac{4}{13}\right)$$
$$l_1l_2 + m_1m_2 + n_1n_2= \frac{36}{169} + \frac{12}{169} – \frac{48}{169} = 0$$
Hence, the lines are perpendicular.
Final Result
Since all three pairs satisfy
$$l_1l_2 + m_1m_2 + n_1n_2 = 0$$
the given lines are mutually perpendicular.
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NCERT Question 2 : Show that the line through the points $(1, -1, 2) $ and $(3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.
Solution
Let AB be the line joining the points $(1, -1, 2) $ and $(3, 4, -2)$, and CD be the line joining the points $(0, 3, 2)$ and $(3, 5, 6)$.
Direction ratios of the line $AB$ joining the points $(1, -1, 2) $ and $(3, 4, -2)$ :
$$a_1 = 3 – 1, b_1 = 4 – (-1), c_1 = -2 – 2$$
$$a_1 = 2, b_1 = 5, c_1 = -4$$
Direction ratios of the line $CD$ joining the points $(0, 3, 2)$ and $(3, 5, 6)$ :
$$a_2 = 3 – 0, b_2 = 5 – 3, c_2 = 6 – 2$$
$$a_2 = 3, b_2 = 2, c_2 = 4$$
To check perpendicularity:
$$AB \perp CD \Rightarrow a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$$
$$a_1 a_2 + b_1 b_2 + c_1 c_2 = 2 \cdot 3 + 5 \cdot 2 + (-4) \cdot 4$$
$$a_1 a_2 + b_1 b_2 + c_1 c_2= 6 + 10 – 16 = 0$$
Hence, AB and CD are perpendicular to each other.
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NCERT Question 3 : 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).
Solution
Let AB be the line through the points (4, 7, 8) and (2, 3, 4), and CD be the line through the points (-1, -2, 1) and (1, 2, 5).
Direction ratios of the line $AB$ joining the points (4, 7, 8) and (2, 3, 4) :
$$a_1 = 2 – 4, b_1 = 3 – 7, c_1 = 4 – 8$$
$$a_1 = -2, b_1 = -4, c_1 = -4$$
Direction ratios of the line $CD$ joining the points (-1, -2, 1) and (1, 2, 5) :
$$a_2 = 1 – (-1), b_2 = 2 – (-2), c_2 = 5 – 1$$
$$a_2 = 2, b_2 = 4, c_2 = 4$$
To check parallelism:
$$AB \parallel CD \Rightarrow \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$
$$\frac{a_1}{a_2} = \frac{-2}{2} = -1$$
$$\frac{b_1}{b_2} = \frac{-4}{4} = -1$$
$$\frac{c_1}{c_2} = \frac{-4}{4} = -1$$
Since all ratios are equal, the lines are parallel.
Hence, AB is parallel to CD.
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NCERT Question 4 : Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $(3\hat{i} + 2\hat{j} – 2\hat{k})$.
Solution
The line passes through point $A(1, 2, 3)$.
The position vector of point $A$ is:
$$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$$
The given direction vector is:
$$\vec{b} = 3\hat{i} + 2\hat{j} – 2\hat{k}$$
A line through point $A$ and parallel to $\vec{b}$ is given by:
$$\vec{r} = \vec{a} + \lambda \vec{b}$$
Thus,
$$\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(3\hat{i} + 2\hat{j} – 2\hat{k})$$
This is the required equation of the line.
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NCERT Question 5 : Find the equation of the line in vector and Cartesian form that passes through the point with position vector $(2\hat{i} – \hat{j} + 4\hat{k})$ and is in the direction $(\hat{i} + 2\hat{j} – \hat{k})$.
Solution
The position vector of the given point is:
$$\vec{a} = 2\hat{i} – \hat{j} + 4\hat{k}$$
The direction vector of the line is:
$$\vec{b} = \hat{i} + 2\hat{j} – \hat{k}$$
A line passing through point $A$ and parallel to $\vec{b}$ is:
$$\vec{r} = \vec{a} + \lambda \vec{b}$$
Substituting the vectors:
$$\vec{r} = 2\hat{i} – \hat{j} + 4\hat{k} + \lambda(\hat{i} + 2\hat{j} – \hat{k})$$
This is the required vector equation of the line.
Let the position vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.
Then,
$$x\hat{i} + y\hat{j} + z\hat{k} = (2 + \lambda)\hat{i} + ( -1 + 2\lambda )\hat{j} + (4 – \lambda)\hat{k}$$
Equating components:
$$
x = 2 + \lambda,\quad
y = -1 + 2\lambda,\quad
z = 4 – \lambda
$$
Eliminating $\lambda$:
$$\frac{x – 2}{1} = \frac{y + 1}{2} = \frac{z – 4}{-1}$$
This is the Cartesian equation of the required line.
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NCERT Question 6 : Find the Cartesian equation of the line which passes through the point $(-2, 4, -5)$ and is parallel to $$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$$
Solution
It is given that the line passes through the point $(-2, 4, -5)$ and is parallel to
$$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$$
Direction ratios of the line
$$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$$
are $3, 5, 6$.
Required line is parallel to
$$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$$
Therefore, its direction ratios are $3k, 5k, 6k$ (where $k\neq 0$).
It is known that the equation of the line through the point $(x_1, y_1, z_1)$ with direction ratios $a, b, c$ is
$$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$$
Hence, equation of the required line passes through the point $(-2, 4, -5)$ and parallel to given line whose direction ratios are $3k, 5k, 6k$ is
$$\frac{x+2}{3k}=\frac{y-4}{5k}=\frac{z+5}{6k}$$
Cancelling $k$ from all denominators, we get
$$\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$$
Final Result
$$\boxed{\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}}$$
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NCERT Question 7 : Write the vector form of the line whose Cartesian equation is $$\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$$
Solution
The given line passes through the point $(5, -4, 6)$. The position vector of this point is
$$\vec{a} = 5\hat{i} – 4\hat{j} + 6\hat{k}$$
The direction ratios of the line are $3, 7, 2$, which corresponds to the direction vector
$$\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$$
The vector form of a line passing through point $\vec{a}$ and in the direction of $\vec{b}$ is
$$\vec{r} = \vec{a} + \lambda \vec{b}, ;\lambda \in \mathbb{R}$$
Substituting the values of $\vec{a}$ and $\vec{b}$, we get
$$\vec{r} = (5\hat{i} – 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} + 2\hat{k})$$
Final Result
$$\boxed{\vec{r} = (5\hat{i} – 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} + 2\hat{k}),;\lambda \in \mathbb{R}}$$
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NCERT Question 8(i): Find the angle between the lines
$$\vec r=2\hat i-5\hat j+\hat k+\lambda(3\hat i+2\hat j+6\hat k)\quad\text{and}\\[1em]
\vec r=7\hat i-6\hat k+\mu(\hat i+2\hat j+2\hat k).$$
Solution
Direction vectors of the lines are
$$\vec b_1=3\hat i+2\hat j+6\hat k,\qquad \vec b_2=\hat i+2\hat j+2\hat k.$$
Compute the dot product:
$$\vec b_1\cdot\vec b_2=3\cdot1+2\cdot2+6\cdot2=3-4+12=19.$$
Compute the magnitudes:
$$|\vec b_1|=\sqrt{3^2+2^2+6^2}=\sqrt{9+4+36}=\sqrt{49}=7,$$
$$|\vec b_2|=\sqrt{1^2+2^2+2^2}=\sqrt{1+4+4}=\sqrt{9}=3.$$
So
$$\cos\theta=\frac{\vec b_1\cdot\vec b_2}{|\vec b_1||\vec b_2|}=\frac{19}{7\cdot3}=\frac{19}{21}.$$
Therefore the angle between the lines is
$$\theta=\cos^{-1}\left(\frac{19}{21}\right).$$
Final result: $\displaystyle \theta=\cos^{-1}\left(\frac{19}{21}\right)$.
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NCERT Question 8 (ii): Find the angle between the lines
$$\vec r=3\hat i+\hat j-2\hat k+\lambda(\hat i-\hat j-2\hat k)\quad\text{and}\\[1em]
\vec r=2\hat i-\hat j-56\hat k+\mu(3\hat i-5\hat j-4\hat k).$$
Solution
Direction vectors are
$$\vec b_1=\hat i-\hat j-2\hat k,\qquad \vec b_2=3\hat i-5\hat j-4\hat k.$$
Dot product:
$$\vec b_1\cdot\vec b_2=1\cdot 3+(-1)\cdot(-5)+(-2)\cdot(-4)=3+5+8=16.$$
Magnitudes:
$$|\vec b_1|=\sqrt{1^2+(-1)^2+(-2)^2}=\sqrt{6},\\[1em]
|\vec b_2|=\sqrt{3^2+(-5)^2+(-4)^2}=\sqrt{50}=5\sqrt{2}.$$
Product of magnitudes:
$$|\vec b_1||\vec b_2|=\sqrt{6}\sqrt{50}=\sqrt{300}=10\sqrt{3}.$$
Therefore
$$\cos\theta=\frac{\vec b_1\cdot\vec b_2}{|\vec b_1||\vec b_2|}=\frac{16}{10\sqrt{3}}=\frac{8}{5\sqrt{3}}.$$
So the angle is
$$\theta=\cos^{-1}\left(\frac{8}{5\sqrt{3}}\right)\approx \cos^{-1}(0.9230769)\approx 22.62^\circ\approx 0.395\ \text{rad}.$$
Final result: $(\displaystyle \theta=\cos^{-1}\left(\frac{8}{5\sqrt{3}}\right)\approx 22.62^\circ.)$
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NCERT Question 9(i) : Find the angle between the following pair of lines
$$\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3} \quad \text{and} \\[1em] \frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$$
Solution
Let the vectors parallel to the given pair of lines be $\vec b_1$ and $\vec b_2$.
From
$$\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$$
we get
$$\vec b_1 = 2\hat{i}+5\hat{j}-3\hat{k}$$
From
$$\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$$
we get
$$\vec b_2 = -\hat{i}+8\hat{j}+4\hat{k}$$
Now,
$$|\vec b_1|=\sqrt{2^2+5^2+(-3)^2}=\sqrt{38}$$
$$|\vec b_2|=\sqrt{(-1)^2+8^2+4^2}=\sqrt{81}=9$$
Dot product:
$$\vec b_1\cdot\vec b_2=(2)(-1)+(5)(8)+(-3)(4)$$
$$\vec b_1\cdot\vec b_2=-2+40-12=26$$
The angle $\theta$ between the lines is given by:
$$\cos\theta=\frac{\vec b_1\cdot\vec b_2}{|\vec b_1||\vec b_2|}$$
$$\cos\theta=\frac{26}{9\sqrt{38}}$$
Thus,
$$\theta=\cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)$$
Final Result
$$\boxed{\theta=\cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)}$$
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NCERT Question 9 (ii) : Find the angle between the following pair of lines
$$\frac{x}{2}=\frac{y}{2}=\frac{z}{1} \quad \text{and} \\[1em] \frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$$
Solution
Let $\vec b_1$ and $\vec b_2$ be vectors parallel to the given lines.
From
$$\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$$
we get
$$\vec b_1 = 2\hat{i}+2\hat{j}+\hat{k}$$
From
$$\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$$
we get
$$\vec b_2 = 4\hat{i}+\hat{j}+8\hat{k}$$
Now compute magnitudes:
$$|\vec b_1|=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3$$
$$|\vec b_2|=\sqrt{4^2+1^2+8^2}=\sqrt{81}=9$$
Dot product:
$$\vec b_1\cdot\vec b_2=(2)(4)+(2)(1)+(1)(8)=8+2+8=18$$
Angle between the two lines:
$$\cos\theta=\frac{\vec b_1\cdot\vec b_2}{|\vec b_1||\vec b_2|}$$
$$\cos\theta=\frac{18}{3\times 9}=\frac{2}{3}$$
Thus,
$$\theta=\cos^{-1}\left(\frac{2}{3}\right)$$
Final Result
$$\boxed{\theta=\cos^{-1}\left(\frac{2}{3}\right)}$$
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NCERT Question 10 : Find the value of $p$ so that the two lines
$$\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}
\quad and \\[1em]
\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$$
are perpendicular.
Solution
Rewrite each in symmetric form.
For the first line,
$$\frac{1-x}{3}=\frac{7(y-2)}{2p}=\frac{z-3}{2}.$$
This gives direction ratios
$$\vec b_1 = \left(-3,\frac{2p}{7},2\right).$$
For the second line,
$$\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}.$$
This gives direction ratios
$$\vec b_2 = \left(-\frac{3p}{7},1,-5\right).$$
For perpendicular lines,
$$\vec b_1\cdot\vec b_2=0.$$
Compute:
$$
(-3)\left(-\frac{3p}{7}\right)
+\left(\frac{2p}{7}\right)(1)
+2(-5)=0
$$
Simplify:
$$
\frac{9p}{7}+\frac{2p}{7}-10=0
$$
$$
\frac{11p}{7}=10
$$
$$
p=\frac{70}{11}
$$
Final Result
$$\boxed{p=\frac{70}{11}}$$
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