Q.Find the angle between the pair of lines given by $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{i} - 2\hat{j} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$.

Concept: Angle between two lines with direction ratios b1 and b2:

cos(theta) = |b1 . b2| / (|b1| |b2|)

Line 1: direction (3, 5, 4)

Line 2: direction (1, 1, 2)

Dot product:

3(1) + 5(1) + 4(2) = 3 + 5 + 8 = 16

Magnitudes:

|b1| = sqrt(9 + 25 + 16) = sqrt(50) = 5 sqrt(2)

|b2| = sqrt(1 + 1 + 4) = sqrt(6)

cos(theta) = 16 / (5 sqrt2 * sqrt6) = 16 / (5 sqrt12) = 16 / (5 * 2 sqrt3) = 16 / (10 sqrt3) = 8 / (5 sqrt3)

So theta = cos^-1 ( 8 / (5 sqrt 3) ).

(Numerically 8/(5*1.732) = 0.924, giving theta ~ 22.5 degrees - not 30 degrees, and option (B) 3/(2 sqrt2) = 1.06 > 1 is not even a valid cosine.)

ANSWER: C

We have two skew lines in 3D. The line of shortest distance between them is the unique line that is perpendicular to both. So its direction vector must be orthogonal to the direction vectors of both given lines — that is, it must be parallel to the cross product of those two direction vectors.

Step 1: Write the direction vectors of $M_1$ and $M_2$ in standard form.

For $M_1$:

Notice the second term: $\frac{2-y}{1} = \frac{y-2}{-1}$. So the symmetric form is actually:

Thus the direction vector of $M_1$ is:

For $M_2$:

So its direction vector is:

Step 2: Find the direction vector of $L_1$, the line of shortest distance.

$L_1$ is perpendicular to both $M_1$ and $M_2$, so its direction vector $\vec{v}$ is parallel to $\vec{d}_1 \times \vec{d}_2$.

Compute the cross product:

Simplify:

• $i$-component: $(-1)(1) - (1)(2) = -1 - 2 = -3$
• $j$-component: $(1)(1) - (1)(1) = 0$, so $-\hat{j}(0) = 0$
• $k$-component: $(1)(2) - (-1)(1) = 2 + 1 = 3$

Thus:

We can take the direction as $\langle 1, 0, -1 \rangle$ (since scaling doesn't change direction).

Step 3: Identify the direction of $L_2$.

$L_2$ is parallel to $\vec{b} = \hat{j} + \hat{k} = \langle 0, 1, 1 \rangle$.

Step 4: Find the acute angle between $L_1$ and $L_2$.

The angle $\theta$ between two lines is the acute angle between their direction vectors. Use the dot product formula:

Here $\vec{v} = \langle 1, 0, -1 \rangle$, $\vec{b} = \langle 0, 1, 1 \rangle$.

Dot product:

Magnitudes:

Thus:

So $\theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ$.

Step 5: Match with the options.

$60^\circ$ corresponds to option (D).

Concept & Intuition

The angle between two lines in space is defined as the acute angle between their direction vectors. If the direction vectors are $\mathbf{a}$ and $\mathbf{b}$, then

When the dot product is zero, the lines are perpendicular (angle $\pi/2$). The trick here is to correctly extract the direction vectors from the given equations — especially the first line, which is given in parametric form with parameter $k$.

Step-by-step solution

1. Find the direction vector of the first line. The line is given as

This is parametric form with parameter $k$. The coefficients of $k$ give the direction vector:

(The constant terms $(1,-1,3)$ give a point on the line, but we only need the direction.)

2. Find the direction vector of the second line. The line is given in symmetric form:

The denominators are the components of the direction vector:

3. Compute the dot product.

4. Interpret the result.

   A dot product of zero means the direction vectors are orthogonal. Therefore the angle between the lines is $90^\circ$, i.e., $\pi/2$ radians.

5. Check the options.

   • (A) $\cos^{-1}(4/9)$ — not zero.
   • (B) $\sin^{-1}(4/3)$ — invalid, since $\sin^{-1}$ argument cannot exceed 1.
   • (C) $\sin^{-1}(\sqrt{5}/3)$ — not zero.
   • (D) $\pi/2$ — matches our result.

From $l+2m+3n=0$, $l=-2m-3n$. Substitute into $3lm-4ln+mn=0$:

Take $n=1$: $m=\sqrt2$ gives $l=-2\sqrt2-3$; $m=-\sqrt2$ gives $l=2\sqrt2-3$.

Dot product of the two triples $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$:

Hence $\cos\theta=0$ and $\theta=\dfrac{\pi}{2}$.

Concept: write each symmetric-form line in the standard x/a = y/b = z/c form to read off direction ratios, then use cos(theta) proportional to the dot product.

Line 1: 2x = 3y = -z. Put each equal to t: x = t/2, y = t/3, z = -t.

Direction ratios (1/2, 1/3, -1); multiply by 6: (3, 2, -6).

Line 2: 6x = -y = -4z = s: x = s/6, y = -s, z = -s/4.

Direction ratios (1/6, -1, -1/4); multiply by 12: (2, -12, -3).

Dot product = (3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0.

A zero dot product means the direction vectors are perpendicular, so the angle is 90 degrees.

ANSWER: C

We are given two conditions on the direction cosines $(l, m, n)$ of a line:

1. $l + m + n = 0$
2. $l^2 = m^2 + n^2$

These conditions define not one line, but two distinct lines (since the equations are symmetric and quadratic). Our goal is to find the angle between these two lines.

Concept and intuition:

Direction cosines satisfy $l^2 + m^2 + n^2 = 1$. The given equations are constraints that pick out specific directions. By solving the system, we’ll get two possible sets of $(l, m, n)$ (up to sign, which doesn’t affect the line direction). The angle between the lines is then found from the dot product of their direction vectors.

1. Use the condition $l + m + n = 0$ to eliminate one variable. From $l = - (m + n)$. Substitute into $l^2 = m^2 + n^2$:

Expand:

Simplify:

So either $m = 0$ or $n = 0$.

2. Case 1: $m = 0$.

   Then $l + 0 + n = 0 \Rightarrow l = -n$.

   Also $l^2 = 0^2 + n^2 = n^2$, which is automatically true since $l = -n$ gives $l^2 = n^2$.

   So one direction vector (up to scaling) is $(l, m, n) = (-n, 0, n) = n(-1, 0, 1)$.

   A convenient direction vector for the first line is $\mathbf{v}_1 = (-1, 0, 1)$.

3. Case 2: $n = 0$.

   Then $l + m + 0 = 0 \Rightarrow l = -m$.

   And $l^2 = m^2 + 0^2 = m^2$, again automatically satisfied.

   So the direction vector is $(l, m, n) = (-m, m, 0) = m(-1, 1, 0)$.

   A convenient direction vector for the second line is $\mathbf{v}_2 = (-1, 1, 0)$.

4. Find the angle between the two lines.

   The angle $\theta$ between lines with direction vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ satisfies:

Compute dot product:

Magnitudes:

So:

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

The key idea is that the acute angle $\theta$ between two lines with slopes $m_1$ and $m_2$ satisfies

Here $\theta = 45^\circ$, so $\tan 45^\circ = 1$. This gives an equation we can solve for the unknown slope.

1. Set up the formula Let $m_1 = \frac14$ and let the unknown slope be $m_2$. Then

Since $\tan 45^\circ = 1$, we have

2. Remove the absolute value This gives two cases:

3. Solve the first case

Multiply through by 4: $4m_2 - 1 = 4 + m_2$, so $3m_2 = 5$, giving

4. Solve the second case

Multiply by 4: $4m_2 - 1 = -4 - m_2$, so $5m_2 = -3$, giving

5. Check the options The given choices are $\frac54$, $\frac53$, $\frac45$, $\frac35$. Only $\frac53$ appears.

Concept: angle between two lines: tan(theta) = |(m1 - m2)/(1 + m1 m2)|.

The given line 3x - y = -5 has slope 3. Let the required slope be m, with theta = 45 degrees, so tan(theta) = 1:

|(m - 3)/(1 + 3m)| = 1.

Case 1: (m - 3)/(1 + 3m) = 1 => m - 3 = 1 + 3m => -2m = 4 => m = -2.

Case 2: (m - 3)/(1 + 3m) = -1 => m - 3 = -1 - 3m => 4m = 2 => m = 1/2.

So the slopes are -2 and 1/2.

ANSWER: D

Step 1 — Set up the dot product

Direction ratios $(a,b,c)$ and $(b-c,\,c-a,\,a-b)$. Their dot product is

Step 2 — Expand

Every term cancels, regardless of the specific values of $a,b,c$.

Step 3 — Conclude

Since the dot product of the two direction-ratio vectors is identically $0$, the two lines are always perpendicular, i.e. the angle between them is $90°$.

Step 1 — Direction ratios of the first line

$x=k-1,\ y=2k+1,\ z=2k+3$ is a parametric line in $k$. Differentiating (or comparing coefficients of $k$) gives direction ratios

Step 2 — Direction ratios of the second line

$\dfrac{x+1}{2}=\dfrac{y-2}{1}=\dfrac{z-1}{2}$ is already in symmetric form, with direction ratios

Step 3 — Apply the angle formula

Step 4 — Conclude

Concept: the pair of angle bisectors of ax^2 + 2hxy + by^2 = 0 is (x^2 - y^2)/(a - b) = xy/h.

First pair: x^2 - 2mxy - y^2 = 0, so a = 1, b = -1, 2h = -2m => h = -m.

Its bisector pair: (x^2 - y^2)/(1 - (-1)) = xy/(-m)

=> (x^2 - y^2)/2 = -xy/m

=> m(x^2 - y^2) = -2xy

=> m x^2 + 2xy - m y^2 = 0

=> x^2 + (2/m) xy - y^2 = 0.

This must be the second pair x^2 - 2n xy - y^2 = 0. Comparing the xy coefficients:

2/m = -2n

=> mn = -1.

ANSWER: C