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Three Dimensional Geometry Questions and Answers

31.

The shortest distance between the straight line \(\vec{r} = 3\hat{i} + 12\hat{j} + 5\hat{k} + \lambda(2\hat{i})\), \(\lambda \in \mathbb{R}\) and the x-axis is

Exam Year(s)

KEAM - 2026,

Answer: E

The shortest distance is 13.

The line passes through (3, 12, 5) with direction \(2\hat{i}\), so it is parallel to the x-axis. The distance between parallel lines is the distance from any point of one to the other.

\[ d = \sqrt{y^{2} + z^{2}} = \sqrt{144 + 25} = 13 \]

The x-coordinate plays no part, since moving along the x-axis does not change the distance. 12 is the y-coordinate alone; the distance from the x-axis needs both the y- and z-coordinates.

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32.

Consider the straight line \(\vec{r} = (5\hat{i} + 2\hat{j} - 3\hat{k})\) \(+\, t(4\hat{i} + 6\hat{j} - 7\hat{k}), t \in \mathbb{R}\). Which one of the following points, is a point on the straight line?

Exam Year(s)

KEAM - 2026,

Answer: D

The point is (25,32,-38).

A point on the line has the form \((5 + 4t, 2 + 6t, -3 - 7t)\). For the \(x\)-coordinate to be 25, \(t = 5\), and then

\[ (5 + 20,\ 2 + 30,\ -3 - 35) = (25, 32, -38) \]

Testing the others: (21,24,-31) needs \(t = 4\) from \(x\) but gives \(y = 26\); (17,20,-22) needs \(t = 3\) but gives \(z = -24\); (1,-4,5) needs \(t = -1\) but gives \(z = 4\); (45,66,-36) needs \(t = 10\) but gives \(y = 62\).

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Answer: C

The line is \(\vec{r} = (-\hat{i} + 2\hat{j} - 4\hat{k}) + t(8\hat{i} + \hat{j} + 6\hat{k})\).

Rewrite the given line in standard form \(\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}\):

\[ \begin{aligned} \frac{x + 1}{-4} &= \frac{y + \frac{1}{2}}{-\frac{1}{2}} \\ &= \frac{z - 4}{-3} \end{aligned} \]

So its direction ratios are \((-4, -\frac{1}{2}, -3)\), proportional to \((8, 1, 6)\). A parallel line through \((-1, 2, -4)\) is therefore \(\vec{r} = (-\hat{i} + 2\hat{j} - 4\hat{k}) + t(8\hat{i} + \hat{j} + 6\hat{k})\).

The last option has the right direction but passes through \((-1, 2, -6)\). The others read the denominators without first making the coefficients of \(x\), \(y\), \(z\) equal to 1.

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34.

A straight line passes through the point whose position vector is \(\hat{k}\) . The straight line also passes through the point of intersection of the lines \(\vec{r} = \hat{j} + \lambda\hat{i}, \lambda \in \mathbb{R}\) and \(\vec{r} = \hat{i} + s\hat{j}, s \in \mathbb{R}\) . Then the equation of the straight line is

Exam Year(s)

KEAM - 2026,

Answer: A

The line is \(\vec{r} = \hat{k} + t(\hat{i} + \hat{j} - \hat{k})\).

The two given lines are \((\lambda, 1, 0)\) and \((1, s, 0)\). They meet when \(\lambda = 1\) and \(s = 1\), at the point \((1, 1, 0)\).

The required line joins \((0, 0, 1)\) to \((1, 1, 0)\), so its direction is

\[ (1, 1, 0) - (0, 0, 1) = (1, 1, -1) \]

giving \(\vec{r} = \hat{k} + t(\hat{i} + \hat{j} - \hat{k})\). Every other option's direction, added to \(\hat{k}\), never reaches \((1, 1, 0)\); for instance \(\hat{i} - \hat{j} - \hat{k}\) gives \((1, -1, 0)\) at \(t = 1\).

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35.

The shortest distance between the lines \(\vec{r} = -\hat{i} + t\hat{k}, t \in \mathbb{R}\) and \(\vec{r} = -\hat{j} + s\hat{i}, s \in \mathbb{R}\) , is

Exam Year(s)

KEAM - 2026,

Answer: E

The shortest distance is 1.

Line 1 passes through \(A(-1, 0, 0)\) with direction \(\vec{d_1} = \hat{k}\); line 2 passes through \(B(0, -1, 0)\) with direction \(\vec{d_2} = \hat{i}\).

\[ \begin{aligned} \vec{d_1} \times \vec{d_2} &= \hat{k} \times \hat{i} = \hat{j} \\ \overrightarrow{AB} &= \hat{i} - \hat{j} \\ d &= \frac{|\overrightarrow{AB} \cdot \hat{j}|}{|\hat{j}|} = |-1| = 1 \end{aligned} \]

Geometrically, line 1 is vertical through \(x = -1, y = 0\) and line 2 lies along \(y = -1\) in the plane \(z = 0\); they are skew and 1 unit apart in the \(y\)-direction.

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36.

The straight line \(\vec{r} = (2 - 3t)(\hat{i} + \hat{j})\) + \((6t + 1)(\hat{j} + \hat{k})\) + \((12t - 11)(\hat{k} + \hat{i}), t \in \mathbb{R}\) , is parallel to the vector

Exam Year(s)

KEAM - 2026,

Answer: B

The line is parallel to \(3\hat{i} + \hat{j} + 6\hat{k}\).

Collect the terms in \(t\); their coefficient vector is the direction of the line:

\[ \begin{aligned} &-3t(\hat{i} + \hat{j}) + 6t(\hat{j} + \hat{k}) + 12t(\hat{k} + \hat{i}) \\ &= t(9\hat{i} + 3\hat{j} + 18\hat{k}) \\ &= 3t(3\hat{i} + \hat{j} + 6\hat{k}) \end{aligned} \]

Any multiple of \(3\hat{i} + \hat{j} + 6\hat{k}\) is parallel to the line. The other options flip a sign or change a component, so they are not multiples of it.

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37.

If α , β , γ are the angles made by the straight line \(\dfrac{x - 5}{2} = \dfrac{y + 1}{3} = \dfrac{z - 2}{\sqrt{7}}\) with the x-axis, y-axis and z-axis respectively, then \(\cos\alpha, \cos\beta, \cos\gamma\) are respectively,

Exam Year(s)

KEAM - 2026,

Answer: E

The direction cosines are \(\dfrac{1}{\sqrt{5}}, \dfrac{3}{2\sqrt{5}}, \dfrac{\sqrt{7}}{2\sqrt{5}}\).

The denominators of the symmetric form give the direction ratios \(2, 3, \sqrt{7}\). Divide each by their length:

\[ \begin{aligned} \sqrt{4 + 9 + 7} &= \sqrt{20} = 2\sqrt{5} \\ \cos\alpha &= \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}} \end{aligned} \]

Similarly \(\cos\beta = \frac{3}{2\sqrt{5}}\) and \(\cos\gamma = \frac{\sqrt{7}}{2\sqrt{5}}\). Check: \(\frac{4 + 9 + 7}{20} = 1\). The options with a minus sign or with 7 or \(3\sqrt{7}\) in place of 3 or \(\sqrt{7}\) fail this check.

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38.

If the lines \(\frac{2x - 1}{2} = \frac{3 - y}{1} = \frac{z - 1}{3}\) and \(\frac{x + 3}{2} = \frac{y + 2}{5} = \frac{z + 1}{a}\) are perpendicular to each other, then the value of \(a\) is equal to

Exam Year(s)

KEAM - 2026,

Answer: E

\(a\) is 1.

Put the first line in standard form so that each numerator is just \(x\), \(y\) or \(z\) minus a constant:

\[ \begin{aligned} \frac{x - \frac{1}{2}}{1} &= \frac{y - 3}{-1} \\ &= \frac{z - 1}{3} \end{aligned} \]

Its direction ratios are \((1, -1, 3)\); those of the second line are \((2, 5, a)\). Perpendicular lines have zero dot product:

\[ \begin{aligned} 1(2) + (-1)(5) + 3a &= 0 \\ 3a &= 3, \quad a = 1 \end{aligned} \]

Reading the direction ratios directly from the printed denominators, (2, 1, 3), gives \(4 + 5 + 3a = 0\) and a negative value, which is not offered.

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Answer: A

The key gives \(\frac{x - a}{a} = \frac{y - b}{0} = \frac{z - c}{0}\).

A line parallel to the x-axis has direction ratios proportional to \((1, 0, 0)\): it changes only in x. The line through \((a, b, c)\) is usually written \(\frac{x - a}{1} = \frac{y - b}{0} = \frac{z - c}{0}\), and the keyed option has direction ratios \((a, 0, 0)\), which are proportional to \((1, 0, 0)\) as long as \(a \ne 0\).

Every other option has a non-zero y or z direction ratio, so those lines are not parallel to the x-axis; option E, for example, is parallel to the y-axis. (A zero denominator here only means that coordinate stays fixed.)

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40.

The equation of a line passing through the point (1,-2,3) and equally inclined to the axes are

Exam Year(s)

KEAM - 2026,

Answer: E

The line is \(x - 1 = y + 2 = z - 3\).

A line equally inclined to the three axes has equal direction cosines, so its direction ratios can be taken as \((1, 1, 1)\). Through \((1, -2, 3)\) the line is

\[ \frac{x - 1}{1} = \frac{y + 2}{1} = \frac{z - 3}{1} \]

The other options use the wrong signs for the point: \(y - 2\) would pass through \(y = 2\), and \(z + 3\) through \(z = -3\).

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Tags: Three Dimensional Geometry Questions and Answers