4 Marks

Question 14 Marks

Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.

Answer

Suppose the points are A(2, 3, 4), B(-1, -2, 1) and C(5, 8, 7).

We know that the direction ratios of the line joining the points (x₁, y₁, z₁) and (x₂, y₂, z₂) are x₂- x_1,y₂- y_1,z₂- z_1.

The direction ratios of AB are (-1 - 2), (-2 - 3), (1 - 4),

i.e. -3, -5, -3.

The direction ratios of BC are (5 - (-1)), (8 - (-2)), (7 - 1),

i.e. 6, 10, 6.

It can be seen that the direction ratios of BC are -2 times that of AB, i.e. they are proportional. Therefore, AB is parallel to BC.

Since point B is common in both AB and BC, points A, B, and C are collinear.

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Question 24 Marks

Find the angle between the lines whose direction cosines are given by the equations:
l + m +n = 0 and l² + m² + n² = 0

Answer

Given that, l + m + n = 0

l² + m² + n² = 0

From equation (1),

l = -(m + n)

Put the value of l in equation (2),

[-(m + n)]² + m² - n² = 0

(m + n)² + m² - n² = 0

m²+ n² + 2mn + m² - n² = 0

2m² + 2mn = 0

2m(m + n) = 0

m = 0, m + n = 0

m = -n and m = 0

Put the value of m = -n in equation (1)

l = -(m +n)

= -(0 + n)

l = -n

Thus, the direction ratios are proportional to

0, -n, n and -n, 0, n

⇒ 0, -1 1 and -1, 0, 1

So, vectors parallel to these lines are

$\vec{a}=0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{b}=-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}}$ respectively.

Let, $\theta$ be the angle between the $\vec{a}$ and $\vec{b}$

So, $\cos\theta=\frac{\vec{a}\times\vec{b}}{\big|\vec{a}\big|\big|\vec{b}\big|}$

$\vec{a}=0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{b}=-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}}$ respectively.

$\cos\theta=\frac{(0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\times(-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}})}{\sqrt{0^2+(-1)^2+(1)^2}\sqrt{(-1)^2+(0)^2+(1)^2}}$

$=\frac{(0)(-1)+(-1)(0)+(1)(1)}{\sqrt{1+1}\sqrt{1+1}}$

$=\frac{0+0+1}{\sqrt{2}\times\sqrt{2}}$

$=\frac{1}{2}$

$\theta=\cos^{-1}\Big(\frac{1}{2}\Big)$

$\theta=\frac{\pi}{3}$

So, angle between the lines $=\frac{\pi}{3}$.

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Question 34 Marks

Using direction ratios show that the points A(2, 3, -4) B(1, - 2, 3) and C(3, 8, -11) are collinear.

Answer

Here A(2, 3, -4), B(1, -2, 3) and C(3, 8, -11).
Direction ratios of AB = (1 - 2, -2 - 3, 3 + 4) = (-1, -5, 7)
Direction ratios of BC = (3 - 1, 8 + 2, -11 - 3) = (2, 10, -14)
Here, the respective direction consines of AB and AC,
$\frac{-1}{2}=\frac{-5}{10}=\frac{7}{-14}$ are proportional.
Also, Bis the common point between the two lines,
$\therefore$ The points A(2, 3, -4) B(1, -2, 3) and C(3, 8, -11) are collinear.

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Question 44 Marks

Find the angle between the lines whose direction cosines are given by the equations:
2l - m + 2n = 0 and mn + nl + lm = 0

Answer

Given that,
2l - m + 2n = 0 .....(1)
mn + nl + lm = 0 .....(2)
From equation (1),
2l - m + 2n = 0
m = 2l + 2n
Put the value of m in equation (2),
mn + nl + lm = 0
(2l + 2n) n + nl + l(2l + 2n) = 0
2ln + 2n² +nl +2l² + 2ln = 0
2l² + 5ln + 2n² = 0
2l² + 4ln + ln + 2ln² = 0
2l (l + 2n) + n(l + 2n) = 0
(1 + 2n) (2l = n) = 0
l + 2n = 0 or 2l + n = 0
l = -2n or $\text{l}=-\frac{\text{n}}{2}$
Put the value of l = -2n in equation (1)
2l - m + 2n = 0
2 (-2n) - m + 2n = 0
-4n - m + 2n = 0
-2n - m = 0
-2n = m
m = -2n
Again, put the value of $\text{l}=-\frac{1}{2}$ in equation (1)
2l - m + 2n = 0
$2\Big(-\frac{1}{2}\text{n}\Big)-\text{m}+2\text{n}=0$
-n - m + 2n = 0
-m + n = 0
-m = -n
m = n
So, direction cosines of the lines are given by,
-2n, -2n, n or $-\frac{1}{2},\text{n},\text{n},\text{n}$
-2, -2, 1 or $-\frac{1}{2},1,1$
So, vectors parallel to these lines
$\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=-\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ respectively.
Let, $\theta$ be the angle between the $\vec{\text{a}}$ and $\vec{\text{b}},$
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{\big(-2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)\times\Big(-\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)}{\sqrt{(-2)^2+(-2)^2+(1)^2}\sqrt{\Big(-\frac{1}{2}^2\Big)+(1)^2+(1)^2}}$
$=\frac{(-2)\big(-\frac{1}{2}\big)+(-2)(1)+(1)(1)}{\sqrt{4+4+1}\sqrt{\frac{1}{4}1+1}}$
$=\frac{1-2+1}{\sqrt{9}\sqrt{\frac{9}{4}}}$
$=\frac{0}{3\times\frac{3}{2}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
So, angle between the lines $=\frac{\pi}{2}$.

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Question 54 Marks

Find the angle between the vectors whose direction cosines are proportional to 2, 3, -6 and 3, -4, 5.

Answer

Let $\vec{\text{a}}$ be a vector with direction ratios 2, 3, -6.
$\Rightarrow\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
Let $\vec{\text{b}}$ be a vector with direction ratios 3, -4, 5.
$\Rightarrow\vec{\text{b}}=3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$
Let $\theta$ be the angle between the given vectors.
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}).(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})}{\big|2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}\big|\big|3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big|}$
$=\frac{6-12-30}{\sqrt{4+9+36}\sqrt{9+16+25}}$
$=\frac{-36}{\sqrt{49}\sqrt{50}}$
$=\frac{-36}{35\sqrt{2}}$
Rationalising the result, we get
$\cos\theta=-\frac{18\sqrt{2}}{35}$
$\therefore\theta=\cos^{-1}\Big(-\frac{18\sqrt{2}}{35}\Big)$
Thus, the angle between the given vectors measures $\cos^{-1}\Big(-\frac{18\sqrt{2}}{35}\Big)$.

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Question 64 Marks

Find the angle between the lines whose direction cosines are given by the equations
l + 2m + 3n = 0 and 3lm - 4ln + mn = 0

Answer

Given that,
l + 2m + 3n = 0 .....(1)
3lm - 4ln + mn = 0 .....(2)
From equation (1),
l + 2m + 3n = 0
l = -2m - 3n
Put the value of l in equation (2),
3lm - 4ln + mn = 0
3(-2m - 3n) m - 4(-2m - 3n) n + mn = 0
-6m²- 9nm + 8mn + 12n² + mn = 0
-6m² + 12n² = 0
m² = 2n²
$\text{m}=\pm\sqrt{2\text{n}^2}$
$\text{m}=\text{n}\sqrt{2}$ or $\text{m}=-\text{n}\sqrt{2}$
Put $\text{m}=\text{n}\sqrt{2}$ in equation (1)
l + 2m + 3n = 0
$\text{l}+2\big(\text{n}\sqrt{2}\big)+3\text{n}=0$
$\text{l}+\text{n}\big(2\sqrt{2}+3\big)=0$
$\text{l}+-\big(2\sqrt{2}+3\big)\text{n}$
Again, $\text{m}=-\sqrt{2\text{n}}$ in equation (1)
l + 2m + 3n = 0
$\text{l}+2\big(-\sqrt{2\text{n}}\big)+3\text{n}=0$
$\text{l}-2\sqrt{2\text{n}}+3\text{n}=0$
$\text{l}+\text{n}\big(-2\sqrt{2\text{n}}+3\big)=0$
$\text{l}=\big(2\sqrt{2\text{n}}-3\big)\text{n}$
Thus, direction cosines of the lines are given by,
$-\big(2\sqrt{2}+3\big)\text{n},\sqrt{2\text{n}},\text{n}$ or $\big(2\sqrt{2}-3\big)\text{n},\sqrt{2\text{n}},\text{n}$
$-\big(2\sqrt{2}+3\big)\text{n},\sqrt{2},1$ or $\big(2\sqrt{2}-3\big)\text{n},-\sqrt{2},1$
So, vectors parallel to these lines are
$\vec{\text{a}}=-\Big(2\sqrt{2}+3\Big)\hat{\text{i}}+\sqrt{2}\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\Big(2\sqrt{2}-3\Big)\hat{\text{i}}-\sqrt{2}\hat{\text{j}}+\hat{\text{k}}$ respectively.
Let, $\theta$ be the angle between the lines,
then,
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\big|\times\big|\vec{\text{b}}\big|}$
$=\frac{-\big(2\sqrt{2}+3\big)\times{\big(2\sqrt{2}-3\big)+\big(\sqrt{2}\big)}\times\big(-\sqrt{2}\big)+(1)(1)}{\sqrt{\big(2\sqrt{2}+3\big)^2+\big(-\sqrt{2}\big)^2+(1)^2}\sqrt{\big(2\sqrt{2}-3\big)^2+\big(-\sqrt{2}\big)^2+(1)^2}}$
$=\frac{-(8-9)-2+1}{\sqrt{8+9+12\sqrt{2}+2+1\sqrt{8+9-12\sqrt{2}+2+1}}}$
$=\frac{-(-1)-2+1}{\sqrt{20+12\sqrt{2}}\sqrt{20-12\sqrt{2}}}$
$=\frac{1-2+1}{\sqrt{20+12\sqrt{2}}\sqrt{20-12\sqrt{2}}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
Angle between the lines $=\frac{\pi}{2}$.

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Question 74 Marks

Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.

Answer

The vectors, represented by these are
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
and $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
Let, $\theta$ be the angle between the lines,
then,
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\Big|\vec{\text{a}}\Big|\Big|\vec{\text{b}}\Big|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{(2)^2+(3)^2+(6)^2}\sqrt{(1)^2+(2)^2+(2)^2}}$
$=\frac{(2)(1)+(3)(2)+(6)(2)}{\sqrt{4+9+36}\sqrt{1+4+4}}$
$=\frac{2+6+12}{\sqrt{49}\sqrt{9}}$
$=\frac{20}{7\times3}$
$\cos\theta=\frac{20}{21}$
$\theta=\cos^{-1}\Big(\frac{20}{21}\Big)$
Angle between the lines $=\cos^{-1}\Big(\frac{20}{21}\Big)$.

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Question 84 Marks

Find the angle between the vectors with direction ratios proportional to 1, -2, 1 and 4, 3, 2.

Answer

Let $\vec{\text{a}}$ be a vector with direction ratios 1, -2, 1.
$\Rightarrow\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
Let $\vec{\text{b}}$ be a vector with direction ratios 4, 3, 2.
$\Rightarrow\vec{\text{b}}=4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
Let $\theta$ be the angle between the given vectors.
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}).(4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\big|\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big|\big|4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\big|}$
$=\frac{4-6+2}{\sqrt{1+4+1}\sqrt{16+9+4}}$
$=\frac{0}{\sqrt{6}\sqrt{29}}$
$=0$
$\therefore\theta=\frac{\pi}{2}$
Thus, the angle between the given vectors measures $\frac{\pi}{2}$.

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Question 94 Marks

Find the angle between the lines whose direction cosines are given by the equations:
2l + 2m - n = 0, mn + ln + lm = 0

Answer

The given equation are,
2l + 2m - n = 0 .....(1)
mn + ln + lm = 0 .....(2)
From (1), We get n = 2l + 2m.
Putting n = 2l + 2m in (2), We get
m(2l + 2m) + l(2l + 2m) + lm = 0
2lm + 2m² + 2l² + 2ml + lm = 0
2ml² + 5lm + 2l² = 0
2m² + 4lm + lm + 2l² = 0
(2m + l) (m + 2l) = 0
$\Rightarrow\text{m}=-\frac{1}{2}$ or $\text{m}=-2\text{l}$
By putting $\text{m}=-\frac{\text{l}}{2}$ in (1) we get n = l
By putting m = 2l in (i) we get n = -2l
So, vector parallel to these lines are
$\vec{\text{a}}=\hat{\text{i}}-\frac{1}{2}\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$ respectively.
If $\theta$ is the angle between the lines, then $\theta$ is also the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$
then,
$\cos\theta=\frac{\vec{\text{a}}\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{1+1-2}{\sqrt{1+\frac{1}{4}+1}\sqrt{1+4+9}}=0$
$\theta=\cos^{-1}(0)=\frac{\pi}{2}$.

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Question 104 Marks

Find the direction cosines of the lines, connected by the relations: l + m + n = 0 and $\frac{2}{\text{m}}+\frac{2}{\text{n}}-\text{mn}=0$.

Answer

Given:
l + m + n = 0 ......(1)
2lm + 2ln + nm = 0 ......(2)
From (1), we get
l = m - n
Substituting l = -m - n in (2), we get
2(-m - n) m + 2(-m - n)n - mn = 0
⇒ -2m² - 2mn - 2mn - 2n² - mn = 0
⇒ 2m² + 2n² + 5mn = 0
⇒ (m + 2n) (2m + n) = 0
$\Rightarrow\text{m}=-2\text{n},-\frac{\text{n}}{2}$
If m = -2n, then from (1), we get l = n.
If $\text{m}=-\frac{\text{n}}{2},$ then from (1), we get $\text{l}=-\frac{\text{n}}{2}.$
Thus, the direction ratios of the two lines are proportional to n, - 2n, n and$-\frac{\text{n}}{2},-\frac{\text{n}}{2},\text{n},\text{i}.\text{e}.1,-2, 1$and $-\frac{1}{2},-\frac{1}{2},1$
Hence, their direction cosines are
$\pm\frac{1}{\sqrt{6}},\pm\frac{-2}{\sqrt{6}},\pm\frac{1}{\sqrt{6}}$
$\pm\frac{-1}{\sqrt{6}},\pm\frac{-1}{\sqrt{6}},\pm\frac{2}{\sqrt{6}}$.

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Question 114 Marks

Find the angle between the lines whose direction ratios are proportional to a, b, c and b - c, c - a, a - b.

Answer

Given, that the direction ratios of lines are proportional to a, b, c and b - c, c - a, a - b.
Let, $\vec{\text{x}}$ and $\vec{\text{y}}$ be the vector parallel to these lines respectively, so
$\vec{\text{x}}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$
And, $\vec{\text{y}}=(\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}$
Let, $\theta$ be the angle between $\vec{\text{x}}$ and $\vec{\text{y}}$, so,
$\cos\theta=\frac{\vec{\text{x}}\times\vec{\text{y}}}{\big|\vec{\text{x}}\big|\big|\vec{\text{y}}\big|}$
$=\frac{(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})[(\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}]}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2+(\text{a}-\text{b})^2}}}$
$=\frac{(\text{a})(\text{b}-\text{c})+\text{b}(\text{c}-\text{a})+\text{c}(\text{a}-\text{b})}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2\sqrt{\text{b}^2+\text{c}^2-2\text{bc}+\text{c}^2+\text{a}^2-2\text{ac}+\text{a}^2+\text{b}^2-2\text{ab}}}}$
$\cos\theta=\frac{\text{ab}-\text{ac}+\text{bc}-\text{ba}+\text{ca}-\text{bc}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
Angle between the lines $=\frac{\pi}{2}$

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Question 124 Marks

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, -4), (-1, 1, 2) and (-5, -5, -2).

Answer

A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2)

The direction ratios of the side AB = (-1 - 3, 1 - 3, 2 + 4)

= (-4, -4 ,6)

Direction cosines of AB will be

$\frac{-4}{\sqrt{(-4)^2+(-4)^2+6^2}},\frac{-4}{\sqrt{(-4)^2+(-4)^2+6^2}},\frac{6}{\sqrt{(-4)^2+(-4)^2+6^2}}$

$=\frac{-4}{\sqrt{68}},\frac{-4}{\sqrt{68}},\frac{6}{\sqrt{68}}$

$=\frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}$

The direction ratios of the side BC = (-5 + 1, -5 - 1, -2 - 2)

= (-4, -6, -4)

Direction cosines of BC will be

$\frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}$

$=\frac{-4}{\sqrt{68}},\frac{-6}{\sqrt{68}},\frac{-4}{\sqrt{68}}$

$=\frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}$

The direction ratios of the side AC = (-5 - 3, -5 - 5, -2 + 4)

= (-8, -10, 2)

Direction cosines of AC will be

$\frac{-8}{\sqrt{(-8)^2+(-10)^2+2^2}},\frac{-10}{\sqrt{(-8)^2+(-10)^2+2^2}},\frac{2}{\sqrt{(-8)^2+(-10)^2+2^2}}$

$=\frac{-8}{\sqrt{168}},\frac{-10}{\sqrt{168}},\frac{2}{\sqrt{168}}$

$=\frac{-4}{\sqrt{42}},\frac{-5}{\sqrt{42}},\frac{1}{\sqrt{42}}$.

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Question 134 Marks

If the coordinates of the points A, B, C, are (1, 2, 3), (4, 5, 6), (-4, 3, -6) and (2, 9, 2), then find the angle between AB and CD.

Answer

The given points are A(1, 2, 3), B(4, 5, 6), C(-4, 3, -6) and D(2, 9, 2).

We know that the direction ratios of the line joining the points

(x₁, y₁, z₁) and (x₂, y₂, z₂) are x₂ x₁, y₂ y₁, z₂ z₁. 2

The direction ratios of AB are (4 - 1), (5 - 2), (7 - 3),i.e. (3, 3, 4).

The direction ratios of CD are [2(-4)], (9 - 3), [2(-6)],

i.e. 6, 6, 8.

Let, $\theta$ be the angle between AB and CD.

We have,

$\text{a}_1=3, \text{c}_1=3, \text{c}_1=4$

$\text{a}_2=6, \text{c}_2=6, \text{c}_2=8$

$\therefore\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}_2^2}}$

$=\frac{18+18+32}{\sqrt{9+9+16}\sqrt{36+36+64}}=\frac{68}{68}$

$=1$

$\Rightarrow\theta=0^\circ$

Thus, the angle between AB and CD measures $0^\circ$.

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