Q5. Find the vector and Cartesian equations of the line through (3, −2, −5) and (3, −2, 6).
By the two-point form (NCERT Summary, page 392), a direction is (3 − 3)î + (−2 + 2)ĵ + (6 + 5)k̂ = 11k̂, or simply k̂. So r = 3î − 2ĵ − 5k̂ + λk̂. In Cartesian form (x − 3)/0 = (y + 2)/0 = (z + 5)/1, that is, x = 3 and y = −2.
Q6. Find the line through (2, 0, −1) parallel to the line r = î − ĵ + λ(4î + ĵ − 3k̂), in both forms.
r = 2î − k̂ + μ(4î + ĵ − 3k̂), and (x − 2)/4 = y/1 = (z + 1)/(−3).
Q7. Find the direction cosines of the line (x − 2)/2 = (2y − 5)/(−3) = z + 1.
Write it as (x − 2)/2 = (y − 5/2)/(−3/2) = (z + 1)/1. Scale the direction ratios 2, −3/2, 1 to 4, −3, 2, whose length is √29, so the direction cosines are 4/√29, −3/√29, 2/√29.
Q8. Where does the line (x − 1)/2 = (y + 1)/3 = (z − 2)/(−1) meet the xy-plane?
On the xy-plane z = 0, so z = 2 − λ = 0 gives λ = 2, and the point is (1 + 4, −1 + 6, 0) = (5, 5, 0).
Q9. Find the points of the line r = 2î + ĵ − k̂ + λ(î − 2ĵ + 2k̂) at distance 6 from the point A(2, 1, −1).
A is the point λ = 0, and |î − 2ĵ + 2k̂| = 3, so the points are λ = ±2: (4, −3, 3) and (0, 5, −5).
Q10. Beyond the NCERT text (JEE): find the foot of the perpendicular from P(1, 6, 3) to the line x/1 = (y − 1)/2 = (z − 2)/3.
A general point of the line is Q(λ, 1 + 2λ, 2 + 3λ), so PQ = (λ − 1)î + (2λ − 5)ĵ + (3λ − 1)k̂. Perpendicular to î + 2ĵ + 3k̂ means (λ − 1) + 2(2λ − 5) + 3(3λ − 1) = 0, so 14λ = 14 and λ = 1. The foot is (1, 3, 5).
Q11. Find the vector equation of the line through (1, 2, −4) perpendicular to the lines (x − 8)/3 = (y + 19)/(−16) = (z − 10)/7 and (x − 15)/3 = (y − 29)/8 = (z − 5)/(−5) (NCERT Miscellaneous Exercise on Chapter 11, Q5, page 391).
Here perpendicular means orthogonal directions; the required line need not meet the two given lines. A direction perpendicular to both is (3î − 16ĵ + 7k̂) × (3î + 8ĵ − 5k̂) = 24î + 36ĵ + 72k̂ = 12(2î + 3ĵ + 6k̂). So r = î + 2ĵ − 4k̂ + λ(2î + 3ĵ + 6k̂).