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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता १२

Choose the correct alternative: Iabcabca→× (b→×c→)=(a→×b→)×c→, where abca→,b→,c→ are any three vectors such that bcb→⋅c→≠0 and aba→⋅b→≠0, then aa→ and cc→ are

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प्रश्न

Choose the correct alternative:

I`vec"a" xx  (vec"b" xx vec"c") = (vec"a" xx vec"b") xx vec"c"`, where `vec"a", vec"b", vec"c"` are any three vectors such that `vec"b"*vec"c" ≠ 0` and `vec"a"*vec"b" ≠ 0`, then `vec"a"` and `vec"c"` are

पर्याय

  • perpendicular

  • parallel

  • inclined at an angle `pi/3`

  • inclined at an angle `pi/6`

MCQ
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उत्तर

parallel

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Application of Vectors to 3-dimensional Geometry
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Applications of Vector Algebra - Exercise 6.10 [पृष्ठ २७७]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 6 Applications of Vector Algebra
Exercise 6.10 | Q 13 | पृष्ठ २७७

संबंधित प्रश्‍न

Find the parametric form of vector equation and Cartesian equations of the straight line passing through the point (– 2, 3, 4) and parallel to the straight line `(x - 1)/(-4) = (y + 3)/5 = (8 - z)/6`


Find the acute angle between the following lines.

`vec"r" = (4hat"i" - hat"j") + "t"(hat"i" + 2hat"j" - 2hat"k")`


Find the acute angle between the following lines.

`(x + 4)/3 = (y - 7)/4 = (z + 5)/5, vec"r" = 4hat"k" + "t"(2hat"i" + hat"j" + hat"k")`


Find the acute angle between the following lines.

2x = 3y = – z and 6x = – y = – 4z


The vertices of ΔABC are A(7, 2, 1), 5(6, 0, 3), and C(4, 2, 4). Find ∠ABC


Show that the points (2, 3, 4), (– 1, 4, 5) and (8, 1, 2) are collinear


Show that the lines `(x - 3)/3 = (y - 3)/(-1), z - 1` = 0 and `(x - 6)/2 = (z - 1)/3, y - 2` = 0 intersect. Aslo find the point of intersection


Show that the straight lines x + 1 = 2y = – 12z and x = y + 2 = 6z – 6 are skew and hence find the shortest distance between them


Find the parametric form of vector equation of the straight line passing through (−1, 2, 1) and parallel to the straight line `vec"r" = (2hat"i" + 3hat"j" - hat"k") + "t"(hat"i" - 2hat"j" + hat"k")` and hence find the shortest distance between the lines


Find the foot of the perpendicular drawn: from the point (5, 4, 2) to the line `(x + 1)/2 = (y - 3)/3 = (z - 1)/(-1)`. Also, find the equation of the perpendicular


Choose the correct alternative:

If `[vec"a", vec"b", vec"c"]` = 1, then the value of `(vec"a"*(vec"b" xx vec"c"))/((vec"c" xx vec"a")*vec"b") + (vec"b"*(vec"c" xx vec"a"))/((vec"a" xx vec"b")*vec"c") + (vec"c"*(vec"a" xx vec"b"))/((vec"c" xx vec"b")*vec"a")` is


Choose the correct alternative:

If `vec"a", vec"b", vec"c"` are non-coplanar, non-zero vectors `[vec"a", vec"b", vec"c"]` = 3, then `{[[vec"a" xx vec"b", vec"b" xx vec"c", vec"c" xx vec"a"]]}^2` is equal to


Choose the correct alternative:

The vector equation `vec"r" = (hat"i" - hat"j" - hat"k") + "t"(6hat"i" - hat"k")` represents a straight line passing through the points


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