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Standard XII

Mathematics

Direction Cosines

The direction...

Question

The direction ratios of the line
$x - y + z - 5 = 0 = x - 3 y - 6 a r e$

3, 1, - 2

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2, - 4, 1

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\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{- 2}{\sqrt{14}}

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\frac{2}{\sqrt{41}}, \frac{- 4}{\sqrt{41}}, \frac{1}{\sqrt{41}}

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Solution

The correct option is A $3, 1, - 2$
$x - y + z - 5 = 0 = x - 3 y - 6$

$x - y + z - 5 = 0$ --(I)
$x - 3 y - 6 = 0$ --(II)
The Direction ratios of the line is $(^i -^j +^k) \times (^i - 3^j)$
= $3^i +^j - 2^k$
Hence,the direction ratios is (3,1,-2).
Hence, Option A is correct

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Similar questions

Q. The direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to
(a) 3, 1, −2

(b) 2, −4, 1

(c)

\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{- 2}{\sqrt{14}}

(d)

\frac{2}{\sqrt{41}}, \frac{- 4}{\sqrt{41}}, \frac{1}{\sqrt{41}}

Q. Equation of the plane containing the line

x + 2 y + 3 z - 4 = 0 = 2 x + y - z + 5

and perpendicular to the plane

5 x + 3 y - 6 z + 8 = 0

Q. Find the angle between the following pairs of lines:
(i)

\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} and \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}

(ii)

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} and \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}

(iii)

\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} and \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}

(iv)

\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 and \frac{x + 1}{1} = \frac{2 y - 3}{3} = \frac{z - 5}{2}

(v)

\frac{x - 5}{1} = \frac{2 y + 6}{- 2} = \frac{z - 3}{1} and \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}

(vi)

\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} and \frac{x + 2}{- 1} = \frac{2 y - 8}{4} = \frac{z - 5}{4}

The equation of the plane through the point $(2, - 1, - 3)$ and parallel to the lines $\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z}{- 4} and \frac{x}{2} = \frac{y - 1}{- 3} = \frac{z - 2}{2}$ is

Q. The equation of the tangents to the circle

x^{2} + y^{2} + 6 x + 6 y + 2 = 0

, which is parallel to

3 x + 4 y + 8 = 0

are

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The direction ratios of the linex−y+z−5=0=x−3y−6are

The correct option is A 3,1,−2x−y+z−5=0=x−3y−6x−y+z−5=0--(I)x−3y−6=0--(II)The Direction ratios of the line is (^i−^j+^k)×(^i−3^j) =3^i+^j−2^kHence,the direction ratios is (3,1,-2).Hence, Option A is correct

The direction ratios of the line
$x - y + z - 5 = 0 = x - 3 y - 6 a r e$

The correct option is A $3, 1, - 2$
$x - y + z - 5 = 0 = x - 3 y - 6$

$x - y + z - 5 = 0$ --(I)
$x - 3 y - 6 = 0$ --(II)
The Direction ratios of the line is $(^i -^j +^k) \times (^i - 3^j)$
= $3^i +^j - 2^k$
Hence,the direction ratios is (3,1,-2).
Hence, Option A is correct