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Vector Algebra question

2016 · 9 Apr · Shift 1 · Q34
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Vector Algebra question

2016 · 9 Apr · Shift 1 · Q34

JEE MainMathematicsVector AlgebraMCQ+4 / −1
In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3 i^\widehat ii+j^−k^\widehat j-\widehat kj​−k, −i^-\widehat i−i + 3 j^\widehat jj​ + p k^\widehat kk and 5 i^\widehat ii + q j^−\widehat j-j​− 4 k^\widehat kk, then the point (p, q) lies on a line :
  1. A
    parallel to x-axis.
  2. B
    parallel to y-axis.
  3. C
    making an acute angle with the positive direction of x-axis.
  4. D
    making an obtuse angle with the positive direction of x-axis.
View written solutionFree

Correct answer: C

  1. Let the position vectors be A⃗=(3,1,−1),B⃗=(−1,3,p),C⃗=(5,q,−4).\vec A=(3,1,-1),\quad \vec B=(-1,3,p),\quad \vec C=(5,q,-4).A=(3,1,−1),B=(−1,3,p),C=(5,q,−4).

  2. Since triangle ABCABCABC is right-angled at AAA, the vectors AB→\overrightarrow{AB}AB and AC→\overrightarrow{AC}AC are perpendicular.

    So, AB→⋅AC→=0.\overrightarrow{AB}\cdot \overrightarrow{AC}=0.AB⋅AC=0.

  3. Compute the vectors: AB→=B⃗−A⃗=(−1−3, 3−1, p−(−1))=(−4,2,p+1),\overrightarrow{AB}=\vec B-\vec A=(-1-3,\,3-1,\,p-(-1))=(-4,2,p+1),AB=B−A=(−1−3,3−1,p−(−1))=(−4,2,p+1), AC→=C⃗−A⃗=(5−3, q−1, −4−(−1))=(2,q−1,−3).\overrightarrow{AC}=\vec C-\vec A=(5-3,\,q-1,\,-4-(-1))=(2,q-1,-3).AC=C−A=(5−3,q−1,−4−(−1))=(2,q−1,−3).

  4. Take dot product: (−4)(2)+2(q−1)+(p+1)(−3)=0.(-4)(2)+2(q-1)+(p+1)(-3)=0.(−4)(2)+2(q−1)+(p+1)(−3)=0.

    Simplifying, −8+2q−2−3p−3=0-8+2q-2-3p-3=0−8+2q−2−3p−3=0 2q−3p−13=0.2q-3p-13=0.2q−3p−13=0.

  5. Hence the point (p,q)(p,q)(p,q) satisfies 2q=3p+132q=3p+132q=3p+13 or q=32p+132.q=\frac{3}{2}p+\frac{13}{2}.q=23​p+213​.

  6. This is a straight line in the pqpqpq-plane with slope m=32>0.m=\frac{3}{2}>0.m=23​>0.

    A line with positive slope makes an acute angle with the positive xxx-axis.

  7. Therefore, the correct option is: C\boxed{\text{C}}C​

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