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Question

Mathematics Question on Vectors

Let A , B , C be three points whose position vectors respectively are
a⃗=i^+4j^+3k^\vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}
b⃗=2i^+αj^+4k^,α∈R\vec{b} = 2\hat{i} + α\hat{j} + 4\hat{k}, α ∈ R
c⃗=3i^−2j^+5k^\vec{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}
If α is the smallest positive integer for whicha⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c}
are non collinear, then the length of the median, in ΔABC, through A is:

A

822\frac{\sqrt{82}}{2}

B

622\frac{\sqrt{62}}{2}

C

692\frac{\sqrt{69}}{2}

D

662\frac{\sqrt{66}}{2}

Answer

822\frac{\sqrt{82}}{2}

Explanation

Solution

The correct answer is (A) : 822\frac{\sqrt{82}}{2}
AB→∣∣AC→\stackrel{→}{AB} || \stackrel{→}{AC}
if
12=α−4−6=12\frac{1}{2} = \frac{α-4}{-6} = \frac{1}{2}
⇒ α = 1
a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} are non-collinear for α = 2
( smallest positive integer )
Now , mid point of BC =p(52,0,92)= p ( \frac{5}{2},0,\frac{9}{2} )
Therefore , Length of the median through A = AP
=(52−1)2+(4)2+(92−3)2= \sqrt{(\frac{5}{2} - 1)² + (4)² + (\frac{9}{2} - 3)²}
=94+16+94= \sqrt{\frac{9}{4} + 16 + \frac{9}{4}}
=822= \frac{\sqrt{82}}{2}