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Coplanar vectors or points: set the scalar triple product to zero and find the unknown

This type of question, asked 6 times: 2021 Paper 8, 28 Jul 2022, 29 Jan 2023, 8 Apr 2023, 12 Apr 2023, 15 Apr 2023. All 6 questions below, oldest first.

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  1. Question 1

    2021 Paper 8

    73
    Let OO be the origin. Let OP→=xi^+yj^−k^\overrightarrow{OP}=x\widehat{i}+y\widehat{j}-\widehat{k} and OQ→=−i^+2j^+3xk^,x,y∈R,x>0\overrightarrow{OQ}= -\widehat{i}+ 2\widehat{j}+ 3x\widehat{k},x,y \in R,x > 0, be such that ∣PQ→∣=20|\overrightarrow{PQ}| =\sqrt{20} and the vector OP→\overrightarrow{OP} is perpendicular to OQ→\overrightarrow{OQ}. If OR→=3i^+zj^−7k^,z∈R\overrightarrow{OR}= 3\widehat{i}+z\widehat{j}- 7\widehat{k},z \in R, is coplanar with OP→\overrightarrow{OP} and OQ→\overrightarrow{OQ}, then the value of x2+y2+z2x^{2}+y^{2}+z^{2} is equal to

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  2. Question 2

    28 Jul 2022

    63
    Let the vectors a→=(1+t)i^+(1−t)j^+k^\overrightarrow{a}= (1 + t)\widehat{i}+ (1 - t)\widehat{j}+\widehat{k}, b→=(1−t)i^+(1+t)j^+2k^\overrightarrow{b}= (1 - t)\widehat{i}+ (1 + t)\widehat{j}+ 2\widehat{k} and c→=ti^−tj^+k^,t∈R\overrightarrow{c}= t\widehat{i}- t\widehat{j}+\widehat{k},t \in R be such that for α,β,γ∈R,αa→+βb→+γc→=0→\alpha,\beta,\gamma \in R,\alpha\overrightarrow{a}+ \beta\overrightarrow{b}+ \gamma\overrightarrow{c}=\overrightarrow{0} ⇒α=β=γ=0\Rightarrow \alpha = \beta = \gamma = 0. Then, the set of all values of tt is:

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  3. Question 3

    29 Jan 2023

    88
    Let a→,b→\overrightarrow{a},\overrightarrow{b} and c→\overrightarrow{c} be three non-zero non-coplanar vectors. Let the position vectors of four points A,B,CA,B,C and DD be a→−b→+c→,λa→−3b→+4c→,−a→+2b→−3c→\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c},\lambda\overrightarrow{a}- 3\overrightarrow{b}+ 4\overrightarrow{c}, -\overrightarrow{a}+ 2\overrightarrow{b}- 3\overrightarrow{c} and 2a→−4b→+6c→2\overrightarrow{a}- 4\overrightarrow{b}+ 6\overrightarrow{c} respectively. If AB→,AC→\overrightarrow{AB},\overrightarrow{AC} and AD→\overrightarrow{AD} are coplanar, then λ\lambda is equal to

  4. Question 4

    8 Apr 2023

    164
    Let the vectors u→1=i^+j^+ak^,u→2=i^+bj^+k^{\overrightarrow{u}}_{1}=\widehat{i}+\widehat{j}+ a\widehat{k},{\overrightarrow{u}}_{2}=\widehat{i}+ b\widehat{j}+\widehat{k} and u→3=i^+j^+k^{\overrightarrow{u}}_{3}=\widehat{i}+\widehat{j}+\widehat{k} be coplanar. If the vectors v→1=(a+b)i^+cj^+ck^, v→2=ai^+(b+c)j^+ak^{\overrightarrow{v}}_{1}= (a + b)\widehat{i}+ c\widehat{j}+ c\widehat{k},\ {\overrightarrow{v}}_{2}= a\widehat{i}+ (b + c)\widehat{j}+ a\widehat{k} and v→3=bi^+bj^+(c+a)k^{\overrightarrow{v}}_{3}= b\widehat{i}+ b\widehat{j}+ (c + a)\widehat{k} are also coplanar, then 6(a+6(a + b+cb + c ) is equal to

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  5. Question 5

    12 Apr 2023

    66
    Let a,b,ca,b,c be three distinct real numbers, none equal to one. If the vectors ai^+j^+k^,i^+bj^+k^a\widehat{i}+\widehat{j}+\widehat{k},\widehat{i}+ b\widehat{j}+\widehat{k} and i^+j^+ck^\widehat{i}+\widehat{j}+ c\widehat{k} are coplanar, then 11−a+11−b+11−c\frac{1}{1 - a}+\frac{1}{1 - b}+\frac{1}{1 - c} is equal to

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  6. Question 6

    15 Apr 2023

    79
    Let SS be the set of all (λ,μ)(\lambda,\mu) for which the vectors λi^−j^+k^,i^+2j^+μk^\lambda\widehat{i}-\widehat{j}+\widehat{k},\widehat{i}+ 2\widehat{j}+ \mu\widehat{k} and 3i^−4j^+5k^3\widehat{i}- 4\widehat{j}+ 5\widehat{k}, where λ−μ=5\lambda - \mu = 5, are coplanar, then ∑(λ,μ)∈S80(λ2+μ2)\sum_{(\lambda,\mu) \in S} 80\left( \lambda^{2}+\mu^{2} \right) is equal to :

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