# Mathematics Questions and Answers – Three Dimensional Geometry – Angle between Two Lines

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Three Dimensional Geometry – Angle between Two Lines”.

1. If L1 and L2 have the direction ratios $$a_1,b_1,c_1 \,and \,a_2,b_2,c_2$$ respectively then what is the angle between the lines?
a) $$θ=tan^{-1}⁡\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |$$
b) $$θ=2tan^{-1}⁡\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |$$
c) $$θ=cos^{-1}⁡\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |$$
d) $$θ=2 \,cos^{-1}⁡\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |$$

Explanation: If L1 and L2 have the direction ratios $$a_1,b_1,c_1 \,and \,a_2,b_2,c_2$$ respectively then the angle between the lines is given by
$$cos⁡θ=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |$$
$$θ=cos^{-1}⁡\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |$$

2. Find the angle between the lines.
$$\frac{x+2}{1}=\frac{y+5}{6}=\frac{z-3}{2}$$
$$\frac{x-4}{5}=\frac{y-3}{-2}=\frac{z+3}{1}$$
a) $$cos^{-1}\frac{⁡5}{\sqrt{1230}}$$
b) $$cos^{-1}⁡\frac{⁡3}{\sqrt{3120}}$$
c) $$cos^{-1}⁡\frac{⁡7}{\sqrt{2310}}$$
d) $$cos^{-1}\frac{⁡⁡48}{\sqrt{1230}}$$

Explanation: We know that, the angle between two lines is given by the formula
cos⁡θ=$$\left |\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |$$
cos⁡θ=$$\left |\frac{1(5)+6(-2)+2(1)}{\sqrt{1^2+6^2+2^2).√(5^2+(-2)^2+1^2}}\right |$$
=$$\left |\frac{-5}{\sqrt{41}.\sqrt{30}} \right |=\frac{5}{\sqrt{1230}}$$
∴$$θ=cos^{-1}\frac{5}{\sqrt{1230}}$$

3. Find the value of p such that the lines
$$\frac{x-1}{3}=\frac{y+4}{p}=\frac{z-9}{1}$$
$$\frac{x+2}{1}=\frac{y-3}{1}=\frac{z-7}{-2}$$
are at right angles to each other.
a) p=2
b) p=1
c) p=-1
d) p=-2

Explanation: The angle between two lines is given by the equation
$$cos⁡θ=\left |\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |$$
cos⁡90°=$$\left |\frac{3(1)+p(1)+1(-2)}{\sqrt{3^2+p^2+1^2}.\sqrt{1^2+1^2+(-2)^2}}\right |$$
0=$$|\frac{p+1}{\sqrt{10+p^2}.√6}|$$
0=p+1
p=-1

4. Find the angle between the two lines if the equations of the lines are
$$\vec{r}=\hat{i}+\hat{j}+\hat{k}+λ(3\hat{i}-\hat{j}+\hat{k}) \,and \,\vec{r}=4\hat{i}+\hat{j}-2\hat{k}+μ(2\hat{i}+3\hat{j}+\hat{k})$$
a) $$cos^{-1}\frac{⁡4}{\sqrt{14}}$$
b) $$cos^{-1}⁡\frac{7}{\sqrt{154}}$$
c) $$cos^{-1}⁡\frac{4}{154}$$
d) $$cos^{-1}⁡\frac{4}{\sqrt{154}}$$

Explanation: Given that, $$\vec{r}=\hat{i}+\hat{j}+\hat{k}+λ(3\hat{i}-\hat{j}+\hat{k})$$ and $$\vec{r}=4\hat{i}+\hat{j}-2\hat{k}+μ(2\hat{i}+3\hat{j}+\hat{k})$$
We know that, if the equations of two lines are of the form $$\vec{r}=\vec{a_1}+λ\vec{b_1} and \,\vec{r}=\vec{a_2}+μ\vec{b_2}$$ then the angle between the two lines is given by
$$cos⁡θ=\left|\frac{\vec{b_1}.\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right|$$
=$$\left |\frac{(3(2)+(-1)3+1(1)}{\sqrt{3^2+(-1)^2+(1)^2.} \sqrt{2^2+3^2+1^2}}\right |=\frac{4}{\sqrt{11}.\sqrt{14}}=\frac{4}{\sqrt{154}}$$
θ=$$cos^{-1}⁡\frac{4}{\sqrt{154}}$$.

5. If two lines L1 and L2 with direction ratios $$a_1,b_1,c_1 \,and \,a_2,b_2,c_2$$ respectively are perpendicular to each other then
$$a_1 a_2+b_1 b_2+c_1 c_2=0$$
a) True
b) False

Explanation: The given statement is true.
We know that the angle between two lines is given by the formula
cos⁡θ=$$\left |\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |$$
So, if the lines L1 and L2 are perpendicular to each to each other then,
θ=90°
⟹$$a_1 \,a_2+b_1 \,b_2+c_1 \,c_2$$=0

6. Find the value of p such that the lines $$\frac{x+11}{4}=\frac{y+3}{-2}=\frac{z-3}{4} \,and \,\frac{x-3}{p}=\frac{y+12}{2}=\frac{z-3}{-12}$$ are at right angles to each other.
a) p=11
b) p=12
c) p=13
d) p=4

Explanation: We know that, if two lines are perpendicular to each other then,
$$a_1 a_2+b_1 b_2+c_1 c_2=0$$
i.e.4(p)+(-2)2+4(-12)=0
4p-4-48=0
4p=52
p=$$\frac{52}{4}$$=13.

7. If the equations of two lines L1 and L2 are $$\vec{r}=\vec{a_1}+λ\vec{b_1}$$ and $$\vec{r}=\vec{a_2}+μ\vec{b_2}$$, then which of the following is the correct formula for the angle between the two lines?
a) cos⁡θ=$$\left |\frac{\vec{a_1}.\vec{a_2}}{|\vec{b_1}||\vec{a_2}|}\right |$$
b) cos⁡θ=$$\left |\frac{\vec{a_1}.\vec{a_2}}{|\vec{a_1}||\vec{a_2}|}\right |$$
c) cos⁡θ=$$\left |\frac{\vec{b_1}.\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right |$$
d) cos⁡θ=$$\left |\frac{\vec{a_1}.\vec{b_2}}{|\vec{a_1}||\vec{b_2}|}\right |$$

Explanation: Given that the equations of the lines are
$$\vec{r}=\vec{a_1}+λ\vec{b_1} \,and \,\vec{r}=\vec{a_2}+μ\vec{b_2}$$
∴ the angle between the two lines is given by
cos⁡θ=$$\left |\frac{\vec{b_1}.\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right |$$.

8. Find the angle between the lines $$\vec{r}=2\hat{i}+6\hat{j}-\hat{k}+λ(\hat{i}-2\hat{j}+3\hat{k})$$ and $$\vec{r}=4\hat{i}-7\hat{j}+3\hat{k}+μ(5\hat{i}-3\hat{j}+3\hat{k})$$.
a) θ=$$cos^{-1}\frac{⁡20}{\sqrt{602}}$$
b) θ=$$cos^{-1}\frac{⁡20}{\sqrt{682}}$$
c) θ=$$cos^{-1}\frac{⁡8}{\sqrt{602}}$$
d) θ=$$cos^{-1}⁡\frac{14}{\sqrt{598}}$$

Explanation: If two lines have the equations $$\vec{r}=\vec{a_1}+λ\vec{b_1} \,and \,\vec{r}=\vec{a_2}+μ\vec{b_2}$$
Then, the angle between the two lines will be given by
cos⁡θ=$$\left |\frac{\vec{b_1}.\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right |$$
=$$\left |\frac{(\hat{i}-2\hat{j}+3\hat{k}).(5\hat{i}-3\hat{j}+3\hat{k})}{\sqrt{1^2+(-2)^2+(3)^2).√(5^2+(-3)^2+3^2}}\right |$$
=$$\frac{5+6+9}{√14.√43}=\frac{20}{√602}$$
θ=$$cos^{-1}⁡\frac{20}{\sqrt{602}}$$

9. If two lines L1 and L2 are having direction cosines $$l_1,m_1,n_1 \,and \,l_2,m_2,n_2$$ respectively, then what is the angle between the two lines?
a) cot⁡θ=$$\left |l_1 \,l_2+m_1 \,m_2+n_1 \,n_2\right |$$
b) sin⁡θ=$$\left |l_1 \,l_2+m_1 \,n_2+n_1 \,m_2\right |$$
c) tan⁡θ=$$\left |l_1 \,l_2+m_1 \,m_2+n_1 \,n_2\right |$$
d) cos⁡θ=$$\left |l_1 \,l_2+m_1 \,m_2+n_1 \,n_2\right |$$

Explanation: If two lines L1 and L2 are having direction cosines $$l_1,m_1,n_1 \,and \,l_2,m_2,n_2$$ respectively, then the angle between the lines is given by
cos⁡θ=$$\left |l_1 \,l_2+m_1 \,m_2+n_1 \,n_2\right |$$

10. Find the angle between the pair of lines $$\frac{x-3}{5}=\frac{y+7}{3}=\frac{z-2}{2} \,and \,\frac{x+1}{3}=\frac{y-5}{4}=\frac{z+2}{8}$$.
a) $$cos^{-1}⁡\frac{43}{\sqrt{3482}}$$
b) $$cos^{-1}⁡⁡\frac{43}{\sqrt{3382}}$$
c) $$cos^{-1}⁡⁡\frac{85}{\sqrt{3382}}$$
d) $$cos^{-1}⁡⁡\frac{34}{\sqrt{3382}}$$

Explanation: The direction ratios are 5, 3, 2 for L1 and 3, 4, 8 for L2
∴ the angle between the two lines is given by
cos⁡θ=$$\frac{(a_1 a_2+b_1 b_2+c_1 c_2)}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$$
=$$\frac{15+12+16}{\sqrt{5^2+3^2+2^2}.\sqrt{3^2+4^2+8^2}}$$
=$$\frac{43}{\sqrt{38}.\sqrt{89}}=\frac{43}{\sqrt{3382}}$$
θ=$$cos^{-1}⁡\frac{43}{\sqrt{3382}}$$.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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