The distance between the line r -2 -2 -3 k-4 k-and the plane r-5 -k-5A- 10-3 -3B-C- 30-9D-

The correct option is A 10/3√(3)Since (î ĵ+4k̂).(î+5ĵ+k̂)=0, The line is parallel to the plane. The plane in Cartesians system is x + 5y + z = 0 and (2, –2, 3) ...

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Byju's Answer

Standard XII

Mathematics

Equation of Perpendicular from a Point on a Line

The distance ...

Question

The distance between the line $¯ r = 2^i - 2^j - 3^k + λ (^i -^j + 4^k)$
and the plane $¯ r . (^i + 5^j +^k) = 5$

\frac{10}{3 \sqrt{3}}

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\frac{10}{3}

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\frac{10}{9}

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\frac{1}{9}

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Solution

The correct option is A $\frac{10}{3 \sqrt{3}}$
Since $(^i -^j + 4^k) . (^i + 5^j +^k) = 0$ , The line is parallel to the plane.
The plane in Cartesians system is x + 5y + z = 0 and (2, –2, 3) is a point on the line.
$∴$ Required distance = $\frac{| 2 - 10 + 3 - 5 |}{\sqrt{1 + 25 + 1}} = \frac{10}{3 \sqrt{3}}$

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Similar questions

Q. The distance between the line

r = 2^i - 2^j + 3^k + λ (^i -^j + 4^k)

and the plane

r . (^i + 5^j +^k) = 5

Q. The distance between the plane

\to r . (^i + 5^j +^k) = 5

and the line

\to r = 2^i - 2^j + 3^k + λ (^i -^j + 4^k)

Q. The distance between the line

\to r = 2^i - 2^j + 3^k + λ (^i -^j + 4^k)

and the plane

\to r \cdot (^i + 5^j +^k) = 5

is-

Q. The distance (in units) between the parallel lines

\to r =^i + 2^j + 3^k + λ (^i -^j +^k)

and

\to r = 2^i -^j -^k + μ (^i -^j +^k)

Q. The distance between the line

\to r = 2^i - 2^j + 3^k + s (^i -^j + 4^k)

and the plane

\to r . (^i + 5^j +^k) = 5

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The distance between the line ¯r=2^i−2^j−3^k+λ(^i−^j+4^k) and the plane ¯r.(^i+5^j+^k)=5

The correct option is A 103√3Since (^i−^j+4^k).(^i+5^j+^k)=0, The line is parallel to the plane. The plane in Cartesians system is x + 5y + z = 0 and (2, –2, 3) is a point on the line. ∴ Required distance = |2−10+3−5|√1+25+1=103√3

The distance between the line $¯ r = 2^i - 2^j - 3^k + λ (^i -^j + 4^k)$
and the plane $¯ r . (^i + 5^j +^k) = 5$