A line passes through 2-1-3 and is perpendicular to the lines r -k-2 -2 -k- and r -2 -3 k-2 -2 k-Obtain its equation in vector and Certesian form-

Given line are r⃗=(î+ĵ+k̂)+λ (2î+ 2ĵ+k̂) and r⃗=(2î ĵ 3k̂)+μ(î+2ĵ+2k̂) A line perpendicular to the given lines will be in the direction of (2î 2ĵ+k̂)×(î+2ĵ+2k ...

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Byju's Answer

Standard XII

Mathematics

Co-Linearity

A line passes...

Question

A line passes through (2, -1, 3) and is perpendicular to the lines $\to r = (^i +^j +^k) + λ (2^i + - 2^j +^k)$ and $\to r = (2^i -^j - 3^k) + μ (^i + 2^j + 2^k)$ . Obtain its equation in vector and Certesian form.

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Solution

Given line are $\to r = (^i +^j +^k) + λ (2^i + - 2^j +^k) a n d \to r = (2^i -^j - 3^k) + μ (^i + 2^j + 2^k)$

A line perpendicular to the given lines will be in the direction of

$(2^i - 2^j +^k) \times (^i + 2^j + 2^k) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & - 2 & 1 1 & 2 & 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 6^i - 3^j + 6^k o r, \to b = 2^i +^j - 2^k$

Position vector of given point (2, -1, 3) is. $\to a = 2^i -^j + 3^k$

Using $\to r = \to a + λ ¯ b, t h e r e q u i r e d v e c t o r e q u a t o n o f l i n e i s : \to r = 2 h a t i -^j + 3^k + λ (2^i + 2^j - 2^k)$

And, Cartesian form of the line is : $\frac{x - 2}{2} = \frac{y + 1}{1} = \frac{z - 3}{- 2}$

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A line passes through (2, -1, 3) and is perpendicular to the lines →r=(^i+^j+^k)+λ(2^i+−2^j+^k) and →r=(2^i−^j−3^k)+μ(^i+2^j+2^k). Obtain its equation in vector and Certesian form.

A line passes through (2, -1, 3) and is perpendicular to the lines $\to r = (^i +^j +^k) + λ (2^i + - 2^j +^k)$ and $\to r = (2^i -^j - 3^k) + μ (^i + 2^j + 2^k)$ . Obtain its equation in vector and Certesian form.