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Collinear points or vectors: find the unknown

This type of question, asked 3 times: 2021 Paper 26, 29 Jun 2022, 8 Apr 2023. All 3 questions below, oldest first.

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  1. Question 1

    2021 Paper 26

    72
    If vectors a→1=xi^−j^+k^{\overrightarrow{a}}_{1}= x\widehat{i}-\widehat{j}+\widehat{k} and a→2=i^+yj^+zk^{\overrightarrow{a}}_{2}=\widehat{i}+ y\widehat{j}+ z\widehat{k} are collinear, then aa possible unit vector parallel to the vector xi^+yj^+zk^x\widehat{i}+ y\widehat{j}+ z\widehat{k} is:

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  2. Question 2

    29 Jun 2022

    165
    Let A,B,CA,B,C be three points whose position vectors respectively are:
    a→=i^+4j^+3k^\overrightarrow{a}=\widehat{i}+ 4\widehat{j}+ 3\widehat{k}
    b→=2i^+αj^+4k^,α∈R\overrightarrow{b}= 2\widehat{i}+ \alpha\widehat{j}+ 4\widehat{k},\alpha \in R
    c→=3i^−2j^+5k^\overrightarrow{c}= 3\widehat{i}- 2\widehat{j}+ 5\widehat{k}
    If α\alpha is the smallest positive integer for which a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} are non-collinear, then the length of the median, in △ABC\bigtriangleup ABC, through AA is:

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  3. Question 3

    8 Apr 2023

    71
    If the points with vectors αi^+10j^+13k^\alpha\widehat{i}+ 10\widehat{j}+ 13\widehat{k}, 6i^+11j^+11k^,92i^+βj^−8k^6\widehat{i}+ 11\widehat{j}+ 11\widehat{k},\frac{9}{2}\widehat{i}+ \beta\widehat{j}- 8\widehat{k} are collinear, then (19α−6β)2(19\alpha - 6\beta)^{2} is equal to

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