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- If the Lines X − 1 − 3 = Y − 2 − 2 K = Z − 3 2 a N D X − 1 K . . .
Find the vector equation of the line passing through the point (1, −1, 2) and perpendicular to the plane 2x − y + 3z − 5 = 0 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the yz - plane
- Misc 3 - If lines are perpendicular find value of k. x-1 -3 . . .
Misc 3 If the lines (𝑥 − 1) ( − 3) = (𝑦 − 2) 2𝑘 = (𝑧 − 3) 2 and (𝑥 − 1) 3𝑘 = (𝑦 − 1) 1 = (𝑧 − 6) ( − 5) are perpendicular, find the value of k
- If the lines (x-1) (-3)=(y-2) (2k)=(z-3) 2and (x-1) (3k)=(y-1 . . .
To solve the problem, we need to find the value of k such that the two given lines are perpendicular The lines are represented in symmetric form, and we can extract their direction ratios from these equations Step 1: Identify the direction ratios of the lines Step 2: Use the condition for perpendicularity
- If the lines x−1−3=y−2−2k=z−32 and x−1k=y−21=z−35 are . . .
If the lines x−1−3=y−2−2k=z−32 and x−1k=y−21=z−35 are perpendicular, what would be the value of k and the equation of plane containing these lines? See what the community says and unlock a badge Answer: Let's analyze the problem step by step: ### 1 Equations of the Lines Given lines are:
- [FREE] If the lines \frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z . . .
The direction ratios of the line − 3 x − 1 = 2 k y − 2 = 2 z − 3 are (− 3, 2 k, 2) The direction ratios of the line 3 k x − 1 = 1 y − 5 = − 5 z − 6 are (3 k, 1, − 5) For two lines to be perpendicular, the dot product of their direction ratios should be zero Therefore, we have: (− 3) (3 k) + (2 k) (1) + (2) (− 5) = 0
- If the lines (x-1) -3 = (y-2) -2k = (z-3) 2 and (x-1) k = (y . . .
If the lines \(\frac{x-1}{-3}=\frac{y-2}{-2k}=\frac{z-3}{2}\) and \(\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}\) are perpendicular, then find the value of k and hence find the equation of plane containing these lines
- If the lines (1-x 3)= (y-2 2α )= (z-3 2) and (x-1 3α ) =y-1 . . .
Since, lines are perpendicular
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