Question

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Answer 1

Given, equation of line :-

$\bf \frac{x \: - \: 3}{2} = \frac{y \: - \: 3}{1} = \frac{z}{1} = x$

$\bf \therefore \: x = 2r + 3$

$\bf y = r + 3$

$\bf z = r$

So, (2r + 3, r + 3, r) is the direction ratio of two lines that intersect at

$\bf \frac{ \pi}{3}$

with given line & passes through (0,0).

∴ Angle between the line & unknown lines is

$\bf \frac{ \pi}{3}$

Direction ratio of line is (2, 1, 1)

$\bf ∣a∣ \: = \sqrt{ {2}^{2} + {1}^{2} + {1}^{2} } = \sqrt{6}$

$\bf ∣b∣ \: = \sqrt{(2r \: + \: 3) ^{2} } \: + \: (r \: + \: 3)^{2} \: + \: {r}^{2} = \sqrt{ {6r}^{2} \: + \: 18 \: + \: 18 }$

$\bf \cos \frac{ \pi}{3} = \frac{a\: ⋅ \: b}{∣a∣∣b∣}$

$\bf = \frac{1}{2} = \frac{4r \: + \: 6 \: + \: r \: + \: 3 \: + \: r}{ \sqrt{6} \sqrt{ {6r}^{2} \: + \: 18r \: + \: 18 } }$

$\bf = \frac{1}{2} = \frac{6r \: + \: 9}{6 \sqrt{ {r}^{2} \: + \: 3r \: + \: 3 } }$

$\bf \therefore \sqrt{ {r}^{2} \: + \: 3r \: + \: 3} = 3r \: + \: 3$

$\bf = {r}^{2} \: + \: 3r \: + \: 3 \: = {4r}^{2} \: + \: 9 \: + \: 12r$

$\bf = 0 = {3r}^{2} + \: 6 \: + \: 9r$

$\bf =0= {r}^{2} \: + \: 3r \: + \: 2$

$\bf\therefore(r+1)(r+2)=0$

So, direction ratios are (-1 , 1, -2) and (1, 2, -1) lines are

$\bf \frac{x \: - \: 0}{ - 1} = \frac{y \: - \: 0}{1} = \frac{z \: - \: a}{ - 2} \: and \: \frac{x \: - \: 0}{1} = \frac{y \: - \: 0}{2} = \frac{z \: - \: 0}{ - 1} .$

Answer 2

Answer: 104.72%

$\small\sf\underline\blue{Given \: \: equation \: of \: the \: line \: </p><p>is: }$

$\small\tt{ \frac{x - 3}{2} } = \frac{y - 3}{1} = \frac{z}{1} = λ$

$\small\sf{So, direction \: ratios \: of \: the \: line \: are (2, 1, 1)} = a_{1}, b_{1}, \: c_{1},$

$</p><p>\small\sf\underline\green{Any \: point \: on \: the \: given \: line \: </p><p>is \: }P (2λ + 3, λ + 3, λ) ≡$

$a_{2} \: b_{2} \: c_{2}$

$\small\fbox{So, direction ratios of OP are:</p><p>(2λ + 3, λ + 3, λ) ≡ } a_{2} \: b_{2}</p><p> \: c_{2}$

Now, angle between given line is x/3.

Rest answer is in these 2 images.

Please refer to that for your answer.

Thanks....