{"id":3479,"date":"2023-10-05T10:39:59","date_gmt":"2023-10-05T10:39:59","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3479"},"modified":"2023-10-05T10:42:53","modified_gmt":"2023-10-05T10:42:53","slug":"week-5-and-6-ss2-third-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-5-and-6-ss2-third-term-further-mathematics-notes\/","title":{"rendered":"Week 5 and 6 – SS2 Third Term Further Mathematics Notes"},"content":{"rendered":"

WEEK FIVE
\n<\/strong>TOPIC : MECHANICS ; VECTORS OR CROSS PRODUCT ON TWO OR THREE DIMENSION , CROSS PRODUCT OF TWO VECTORS AND APPLICATION OF CROSS PRODUCT
\n<\/strong>Vector Product of two vectors
\nGiven two vectors andwhose directions are inclined at an angle, <\/strong>their vector productis defined as a vector whose magnitude issinand whose directions is perpendicular to both and and <\/strong>also being positive relative to a rotation from themvector and <\/strong>also being positive relative to a rotation from the vectorto the re
\nctor.
\n<\/strong>\u00a0\u00a0\u00a0\u00a0<\/strong>The vector product ofand b <\/strong>is designated
\nb
\n<\/strong>Thus:
\n\u00a0\u00a0\u00a0\u00a0 = x =<\/strong>|| || sin . <\/strong>whereis a unit vector perpendicular to the plane of and .
\n<\/strong> Properties of vector Product
\n<\/strong><\/p>\n

    \n
  1. \n
    x <\/strong>= |b||a| sin(-) 0 <<
    \n\t\t\t\t\t<\/strong><\/div>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= – |a||b| sin (-)
    \n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= – b
    \n<\/strong>\u00a0\u00a0\u00a0\u00a0Thus the vector product of two vectors is not commutative .\n<\/li>\n

  2. \n
    (k) x \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= x (k )\n<\/div>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= k (x)
    \n\t\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= k |||| sin <\/strong>)
    \nWhere k is a scalar.\n<\/li>\n

  3. \n
    x ( + c<\/strong>)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= x + <\/strong>x c
    \n<\/strong><\/div>\n

    Distribute law\n<\/li>\n

  4. \n
    x\u00a0\u00a0\u00a0\u00a0= <\/strong>x = \u00a0\u00a0\u00a0\u00a0<\/strong>x\n<\/div>\n

    x \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= – <\/strong>x , <\/strong>x k = = <\/strong>\n\t\t\t\t<\/li>\n<\/ol>\n

      \n
    • \n
      x\n<\/div>\n

      \u00a0\u00a0\u00a0\u00a0 x = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0– <\/strong>x
      \n\t\t\t\t\t<\/strong><\/li>\n<\/ul>\n

        \n
      1. \n
        \u00a0\u00a0\u00a0\u00a0|a x b| = area of parallelogram with sides\n<\/div>\n

        and . <\/strong>\n\t\t\t\t<\/li>\n

      2. \u00a0\u00a0\u00a0\u00a0If x = <\/strong>0 and and b<\/em><\/strong> are non zero vectors, then a<\/em><\/strong> and b<\/em><\/strong> are parallel\n<\/li>\n
      3. \n
        If a<\/em><\/strong> = a1 <\/sub>i<\/em><\/strong>+ a2<\/sub>i<\/em><\/strong> + a3<\/sub>k<\/strong>\n\t\t\t\t<\/div>\n

        \u00a0\u00a0\u00a0\u00a0b <\/em><\/strong>= b1<\/sub> + b2<\/sub>i <\/strong><\/em>+b<\/em>3<\/sub>k<\/strong> then
        \n\"\"\/\"\"\/
        \n\t\t\t\t\u00a0\u00a0\u00a0\u00a0i j k
        \n<\/strong>a\u00a0\u00a0\u00a0\u00a0ba1 <\/sub>a2 <\/sub>a3
        \n<\/sub><\/em><\/strong>\u00a0\u00a0\u00a0\u00a0<\/sub> b1<\/sub> b2 <\/sub>b3
        \n<\/sub><\/em><\/strong>
        \n\u00a0We shall make use of the following important result in determinant of order 2 x 2 and order 3 x 3 defined respectively as follows.
        \n\"\"\/\"\"\/a b
        \nc d = ad \u2013 bc
        \n<\/em>\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/
        \n\t\t\t\t\t<\/em>a b c\u00a0\u00a0\u00a0\u00a0<\/em> e\u00a0\u00a0\u00a0\u00a0 f\u00a0\u00a0\u00a0\u00a0– b d\u00a0\u00a0\u00a0\u00a0f + c \u00a0\u00a0\u00a0\u00a0 d e
        \nd e f\u00a0\u00a0\u00a0\u00a0<\/em>= a h\u00a0\u00a0\u00a0\u00a0 I \u00a0\u00a0\u00a0\u00a0 g\u00a0\u00a0\u00a0\u00a0I\u00a0\u00a0\u00a0\u00a0 g h
        \ng h i
        \n\t\t\t\t\t\t\t<\/sub><\/strong><\/em>
        \n\u00a0The expansion of the determinant of order 3 x 3 is along the first row.
        \n\u00a0\u00a0\u00a0\u00a0Notwithstanding it can be along any other row or any column. <\/p>\n

        \u00a0Example 1
        \n<\/strong>Find the vector of a<\/strong> andb<\/strong> where:
        \na = 4I\u00a0\u00a0\u00a0\u00a0– 3j \u00a0\u00a0\u00a0\u00a0+ 2k, b = i + 2 j \u2013 5k
        \n<\/em><\/strong>Solution
        \n\t\t\t\t\t\t<\/em><\/strong>a = 4 I \u2013 3 j + 2 k
        \n<\/em><\/strong>b = I + 2j \u2013 5k
        \n<\/em><\/strong>\"\"\/\"\"\/
        \n\t\t\t\t\t\t<\/em><\/strong>a <\/em><\/strong>x b = i\u00a0\u00a0\u00a0\u00a0 j\u00a0\u00a0\u00a0\u00a0k <\/em>
        \n\t\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0 4\u00a0\u00a0\u00a0\u00a0 -3\u00a0\u00a0\u00a0\u00a02
        \n\u00a0\u00a0\u00a0\u00a0 1\u00a0\u00a0\u00a0\u00a0 2 -5
        \n\"\"\/\"\"\/\"\"\/
        \n\t\t\t\t\t<\/strong>\"\"\/\"\"\/\"\"\/\"\"\/=i -3\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0 4\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a04\u00a0\u00a0\u00a0\u00a0-3
        \n<\/em><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 –j\u00a0\u00a0\u00a0\u00a0 +k \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
        \n<\/em><\/strong>\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0-5\u00a0\u00a0\u00a0\u00a0 1\u00a0\u00a0\u00a0\u00a0-5\u00a0\u00a0\u00a0\u00a01\u00a0\u00a0\u00a0\u00a02 <\/p>\n

        \u00a0= I (15 -4) \u2013j (-20 -2) + k (8 + 3)
        \n<\/em>= 11i + 20j + 11k
        \n<\/em>
        \n\u00a0If p<\/em>\u00a0\u00a0\u00a0\u00a0=2i \u2013 3j + 4k
        \n<\/em> q \u00a0\u00a0\u00a0\u00a0=5i \u2013 4j \u2013 3k
        \n<\/em>Find :\n<\/li>\n

      4. p x q;
        \n\t\t\t\t\t<\/strong><\/em><\/li>\n
      5. \n
        |p x q|
        \n\t\t\t\t\t\t<\/strong><\/em><\/div>\n

        \"\"\/Solution
        \n<\/strong><\/li>\n

      6. \n
        p x q = i\u00a0\u00a0\u00a0\u00a0 j\u00a0\u00a0\u00a0\u00a0k\n<\/div>\n

        \u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0 -3\u00a0\u00a0\u00a0\u00a04
        \n\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0 4\u00a0\u00a0\u00a0\u00a0-3
        \n\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/=i -3\u00a0\u00a0\u00a0\u00a04\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a04\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0-3
        \n<\/em><\/strong>\u00a0\u00a0\u00a0\u00a0–j\u00a0\u00a0\u00a0\u00a0 +k \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
        \n<\/em><\/strong>\u00a0\u00a0\u00a0\u00a04\u00a0\u00a0\u00a0\u00a0-3\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0-3\u00a0\u00a0\u00a0\u00a05\u00a0\u00a0\u00a0\u00a04 <\/p>\n

        \u00a0= i (9 \u2013 8) \u2013j (- 6 -20) + k (8 + 15)
        \n<\/em>=i + 26j + 23 k
        \n<\/em>
        \n\u00a0<\/li>\n

      7. \n
        |p x q| \u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0 |I + 26j + 23k|\n<\/div>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0
        \n`\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0
        \nExample
        \nShow that (a x b)2<\/sup> = a2<\/sup>b2<\/sup> \u2013 (a.b)2<\/sup><\/em>
        \n\t\t\t\t\t<\/strong>Solution
        \n<\/strong>(a x b)2<\/sup>\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0<\/em>(absin)2<\/sup>
        \n\t\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0<\/em>a2<\/sup> b2<\/sup> sin2<\/sup>
        \n\t\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0<\/em>a2<\/sup> b2<\/sup> (1 \u2013 cos2 <\/sup>)
        \n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0<\/em>a2<\/sup> b2<\/sup> – a2<\/sup> b2<\/sup> cos2 <\/sup>
        \n\t\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0<\/em>a2<\/sup> b2 <\/sup>\u2013 (a.b)2<\/sup>
        \n\t\t\t\t\t<\/strong>Hence<\/em>
        \n\t\t\t\t\t<\/strong>(a x b)2\u00a0\u00a0\u00a0\u00a0<\/sup>= \u00a0\u00a0\u00a0\u00a0a2<\/sup> b2<\/sup> \u2013 (a.b)2<\/sup>
        \n\t\t\t\t\t<\/strong>
        \n\u00a0EVALUATION
        \n<\/strong> Given that p = 2i + 3j +4k and q= 5i \u2013 6j +7k find ; (1) p x q ( 2) (p + q ) . ( p-q)<\/p>\n

        \u00a0
        \n\u00a0
        \n\u00a0Application of vector product
        \n<\/strong>Area of a parallelogram Example <\/strong>
        \n\t\t\t\tShow that the area of parallelogram with sides a<\/em><\/strong> andb<\/em><\/strong> is.
        \n\"\"\/Solution
        \n<\/strong>
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0Area of parallelogram
        \n<\/strong>OAC B=\u00a0\u00a0\u00a0\u00a0h\/b
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\/a\/ sin \/<\/strong>b\/
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\/a\/ \/b\/sin
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\/a x b\/
        \nArea of angle
        \nExample
        \n\"\"\/Show that the area of a triangle with sides a and b is |a x b|
        \n\"\"\/Solution
        \n<\/strong>
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0Area of \u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0OAB<\/em> = |b| x h
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0 |b<\/em>||a<\/em>| Sin
        \n\t\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0<\/strong> |a<\/em>||b<\/em>| Sin
        \n\t\t\t\t\t<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0<\/strong> (a x b)
        \n<\/em><\/strong>
        \n\u00a0Example
        \nThe adjacent sides of a parallelogram are
        \n = 2 i<\/strong> \u2013 j \u2013 <\/strong><\/em>6k<\/em><\/strong> and = i<\/em><\/strong> + 3 j \u2013 k <\/em><\/strong>. Find
        \nthe area of the parallelogram. <\/p>\n

        \u00a0\"\"\/Solution
        \n<\/strong>
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0AB \u00a0\u00a0\u00a0\u00a0= \u00a0\u00a0\u00a0\u00a02 i \u2013 j \u2013 <\/em><\/strong>6k
        \n<\/em><\/strong>AC\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0i + 3 j \u2013 k
        \n<\/em><\/strong>\"\"\/\"\"\/Area of parallelogram =\u00a0\u00a0\u00a0\u00a0|AB x AC<\/em>|
        \n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= \u00a0\u00a0\u00a0\u00a0 x
        \n\"\"\/i\u00a0\u00a0\u00a0\u00a0 j k
        \n<\/em> 2\u00a0\u00a0\u00a0\u00a0 -1 -6
        \n 1\u00a0\u00a0\u00a0\u00a0 3 -1\u00a0\u00a0\u00a0\u00a0
        \nx
        \n\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/
        \n\t\t\t\t= i<\/em>\u00a0\u00a0\u00a0\u00a0-1\u00a0\u00a0\u00a0\u00a0-6 –j\u00a0\u00a0\u00a0\u00a0<\/em> 2\u00a0\u00a0\u00a0\u00a0-6 +k\u00a0\u00a0\u00a0\u00a0<\/em>2\u00a0\u00a0\u00a0\u00a0-1
        \n\u00a0\u00a0\u00a0\u00a03\u00a0\u00a0\u00a0\u00a0-1\u00a0\u00a0\u00a0\u00a0 1\u00a0\u00a0\u00a0\u00a0-1\u00a0\u00a0\u00a0\u00a0 1\u00a0\u00a0\u00a0\u00a03\u00a0\u00a0\u00a0\u00a0
        \n= I (1 + 18) \u2013j (-2 + 6) + k (6 + 1)
        \n<\/em>= 19 I \u2013 4 j + 7 k
        \n<\/em>|AB x AC|<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0 |19i – 4j + 7k|
        \n<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/em>=\u00a0\u00a0\u00a0\u00a0
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0
        \nHence
        \nArea of parallelogram = sq. Units <\/p>\n

        \u00a0GENERAL EVALUATION
        \n<\/strong>1) Find the vector product of a= 4i -3j +4k and b = -I + 2j +7k
        \n2) Given that p = 7i + 2j + k and q = 3i \u2013 2j + 4k find ; (i) p x q (ii) | p x q | (iii) the unit vector perpendicular to both p and q
        \n3) Find the sine of the angle between the vectors : a = I \u2013 j + k and b = 8i + 2j + 3k
        \n4) The adjacent sides of a parallelogram are PQ= 4i + 3j + k and PR = -5i + 2j +3k find the area of the parallelogram
        \n5) The position vectors OA, OB and OC are 2i \u2013 3j + 4k , 6i + 4j -8k and 3i + 2j + 5k respectively find (i) vector AB (ii) vector BA (iii) vector BC (iv) AB x BC
        \nReading Assignment: New Further Maths Project 2 page 216 \u2013 222<\/p>\n

        \u00a0WEEKEND ASSIGNMENT
        \n<\/strong>Given that a = I + 2j + k and b = 2i +3j- 5k
        \n1) find ( a x b ) . a a) 0 b) 1 c) 2 d) 3
        \n2) find ( a x b ) . b a) 1 b) 2 c) 0 d) 3
        \nGiven that p = I + 5j + 6k and q = – 2i + j + 3k
        \n3) find p x q a) 15i +11j -11k b) 11i – 15j + 11k c) 11i \u2013 11j + 15k d) 11i- 15j -11k
        \n4) find q x p a) -11i + 15j \u2013 11k b) 11i – 15j + 11k c) 15i \u2013 11j-11k d) 15i+11j+11k
        \n5) Given that a = i \u2013 j+ 3k and b = 6i + 2j \u2013 2k find ( a + b ) . ( a x b ) a) 1 b) 0 c) 2 d) 3<\/p>\n

        \u00a0THEORY<\/strong>
        \n\t\t\t\t1) AB = 4i +3j+5k and AC= 2i-3j+k are two sides of a triangle ABC , find the area of the triangle
        \n2) PQ = 2i+5j+3k and PR = 3i-3j + k are two adjacent sides of a parallelogram, find the area of the parallelogram.<\/p>\n

        \u00a0
        \n\u00a0WEEK SIX
        \n<\/strong>REVIEW OF HALF TERM WORK
        \n<\/strong>
        \n\u00a0
        \n\t\t\t\t\t<\/strong>\u00a0<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

        WEEK FIVE TOPIC : MECHANICS ; VECTORS OR CROSS PRODUCT ON TWO OR THREE DIMENSION…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,275],"tags":[],"class_list":["post-3479","post","type-post","status-publish","format-standard","hentry","category-posts","category-third-term-ss2-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3479","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3479"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3479\/revisions"}],"predecessor-version":[{"id":3480,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3479\/revisions\/3480"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3479"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3479"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3479"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}