Mathematics Crash Course JEE Main>Chapter 28 - Three Dimensional Geometry>Level 1>Q 33

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The two lines $L_{1} : \vec{r} = (\hat{i} + 5 \hat{j} + 5 \hat{k}) + t (4 \hat{i} - 4 \hat{j} + 5 \hat{k})$ and $L_{2} : \vec{r} = (2 \hat{i} + 4 \hat{j} + 5 \hat{k}) + s (8 \hat{i} - 3 \hat{j} + \hat{k})$ are such that

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Important Questions on Three Dimensional Geometry

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Mathematics Crash Course JEE Main>Chapter 28 - Three Dimensional Geometry>Level 1>Q 34

If the lines

\frac{x - k}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}

and

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

are co-planar, then the value of

k

is _______.

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Mathematics Crash Course JEE Main>Chapter 28 - Three Dimensional Geometry>Level 1>Q 35

The length of perpendicular from

(1, 6, 3)

to the line

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

is

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Mathematics Crash Course JEE Main>Chapter 28 - Three Dimensional Geometry>Level 1>Q 36

If

\frac{x - 1}{1} = \frac{y - 2}{m} = \frac{z + 1}{n}

is the equation of the line through

(1, 2, - 1)

and

(- 1, 0, 1)

, then

(l, m, n)

is

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Mathematics Crash Course JEE Main>Chapter 28 - Three Dimensional Geometry>Level 1>Q 37

The equation of line passing through

(3, - 1, 2)

and perpendicular to the lines

\vec{r_{1}} = (\hat{i} + \hat{j} - \hat{k}) + λ (2 \hat{i} - 2 \hat{j} + \hat{k})

and

\vec{r_{2}} = (2 \hat{i} + \hat{j} - 3 \hat{k}) + μ (\hat{i} - 2 \hat{j} + 2 \hat{k})

is

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Mathematics Crash Course JEE Main>Chapter 28 - Three Dimensional Geometry>Level 1>Q 38

The lines

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

and

\frac{x - 4}{2} = \frac{y - 6}{3} = \frac{z - 7}{3}

are coplanar. Their point of intersection is

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Mathematics Crash Course JEE Main>Chapter 28 - Three Dimensional Geometry>Level 1>Q 39

Cosine of the angle between the lines

\vec{r} = 5 \hat{i} - \hat{j} + 4 \hat{k} + λ (\hat{i} + 2 \hat{j} + 2 \hat{k})

and

\vec{r} = 7 \hat{i} + 2 \hat{j} + 2 \hat{k} + μ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})

is

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Mathematics Crash Course JEE Main>Chapter 28 - Three Dimensional Geometry>Level 1>Q 40

The coordinates of a point of the line

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

at distance

4 \sqrt{14}

from the point

(1, - 1, 0)

nearer to the origin are

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Mathematics Crash Course JEE Main>Chapter 28 - Three Dimensional Geometry>Level 1>Q 41

Equation of the line passing through

(1, 1, 1)

and perpendicular to

2 x - 3 y + z = 5

is