{"id":6235,"date":"2021-10-10T21:56:00","date_gmt":"2021-10-10T16:26:00","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6235"},"modified":"2022-01-16T17:04:47","modified_gmt":"2022-01-16T11:34:47","slug":"coplanar-vectors","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/coplanar-vectors\/","title":{"rendered":"Coplanar Vectors – Definition and Example"},"content":{"rendered":"
Here you will learn definition of coplanar vectors with example and test of coplanarity of four points.<\/p>\n
Let’s begin –<\/p>\n
A system of vectors is said to be coplanar, if their supports are parallel to the same plane.<\/p>\n
Note<\/strong> : Two vectors are coplanar.<\/p>\n Theorem 1 (Test of Coplanarity of Three vectors)<\/strong><\/p>\n Let \\(\\vec{a}\\) and \\(\\vec{b}\\) be two given non-zero non-collinear vectors. Then, any vector \\(\\vec{r}\\) coplanar with \\(\\vec{a}\\) and \\(\\vec{b}\\) can be uniquely expressed as \\(\\vec{r}\\) = \\(x\\vec{r}\\) + \\(y\\vec{b}\\), for some scalars x and y.<\/p>\n<\/blockquote>\n Theorem 2<\/strong> <\/p>\n The necessary and sufficient condition for three vectors \\(\\vec{a}\\), \\(\\vec{b}\\) and \\(\\vec{c}\\) to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that \\(l\\vec{a} + m\\vec{b} + n\\vec{c}\\) = \\(\\vec{0}\\).<\/p>\n<\/blockquote>\n Theorem 3<\/strong><\/p>\n If \\(\\vec{a}\\), \\(\\vec{b}\\) and \\(\\vec{c}\\) are three non-zero non-coplanar vector and x, y, z are three scalars, then \\(x\\vec{a} + y\\vec{b} + z\\vec{c}\\) = \\(\\vec{0}\\) \\(\\implies\\) x = y = z = 0.<\/p>\n<\/blockquote>\n Example<\/strong><\/span> : Show that the vectors \\(\\vec{a} – 2\\vec{b} + 3\\vec{c}\\), \\(\\vec{a} – 3\\vec{b} + 5\\vec{c}\\) and \\(-2\\vec{a} + 3\\vec{b} – 4\\vec{c}\\) are coplanar, where \\(\\vec{a}\\), \\(\\vec{b}\\), \\(\\vec{c}\\) are non-coplanar vector.<\/p>\n Solution<\/span><\/strong> : From Theorem 1, <\/p>\n Three vectors are co-planar if one of the given vectors are expressible as a linear combination of the other two. Let<\/p>\n \\(\\vec{a} – 2\\vec{b} + 3\\vec{c}\\) = x\\(\\vec{a} – 3\\vec{b} + 5\\vec{c}\\) + y\\(-2\\vec{a} + 3\\vec{b} – 4\\vec{c}\\) for some scalars x and y.<\/p>\n \\(-2\\vec{a} + 3\\vec{b} – 4\\vec{c}\\) = (x – 2y)\\(\\vec{a}\\) + (-3x + 3y)\\(\\vec{b}\\) + (5x – 4y)\\(\\vec{c}\\)<\/p>\n \\(\\implies\\) 1 = x – 2y, -2 = -3x + 3y and 3 = 5x – 4y<\/p>\n Solving first two of these equation, we get<\/p>\n x = 1\/3, y = -1\/3.<\/p>\n Clearly, these values of x and y satisfy the third equation.<\/p>\n Hence, the given vectors are co-planar.<\/p>\n Four points with position vectors \\(\\vec{a}\\), \\(\\vec{b}\\), \\(\\vec{c}\\) and \\(\\vec{d}\\) are co-planar iff there exist scalars x, y z, u not allzero such that \\(x\\vec{a} + y\\vec{b} + z\\vec{c} + u\\vec{d}\\) = \\(\\vec{0}\\), where x + y + z + u =0.<\/p>\n\n\n\n
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Test of Coplanarity of Four Points<\/h2>\n