Maths · Equation of a line; Skew lines, the shortest distance between them and its equation
are coplanar then
\( \frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-1} \& \frac{x-1}{k}= \) \( \frac{\boldsymbol{y}-\boldsymbol{4}}{\boldsymbol{2}}=\frac{\boldsymbol{z}-\boldsymbol{5}}{\boldsymbol{2}} \) are coplanar then \( \mathbf{k}=? \)
- A. any value
- B. exactly one value
- C. exactly 2 values
- D. exactly 3 values
Step-by-step solution
The condition for coplanarity of two lines is that the scalar triple product of their direction vectors and the vector connecting two points on them is zero. For lines (x-2)/1=(y-3)/1=(z-4)/-1 and (x-1)/k=(y-4)/2=(z-5)/2, with points A(2,3,4) and B(1,4,5) and direction vectors d1=(1,1,-1) and d2=(k,2,2), compute AB = (-1,1,1) and d1×d2 = (4, -2-k, 2-k). Then (d1×d2)·AB = -4 -2k = 0 gives k = -2. Thus only one value satisfies.
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