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Projection of a vector on the sum of two vectors (one with an unknown) is given: find the unknown

This type of question, asked 4 times: 2021 Paper 19, 4 Apr 2024, 8 Apr 2024, 22 Jan 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 19

    83
    If the projection of the vector i^+2j^+k^\widehat{i}+ 2\widehat{j}+\widehat{k} on the sum of the two vectors 2i^+4j^−5k^2\widehat{i}+ 4\widehat{j}- 5\widehat{k} and −λi^+2j^+3k^-\lambda\widehat{i}+ 2\widehat{j}+ 3\widehat{k} is 1 , then λ\lambda is equal to

  2. Question 2

    4 Apr 2024

    169
    Let a→=i^+j^+k^,b→=2i^+4j^−5k^\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\overrightarrow{b}= 2\widehat{i}+ 4\widehat{j}- 5\widehat{k} and c→=xi^+2j^+3k^,x∈R\overrightarrow{c}= x\widehat{i}+ 2\widehat{j}+ 3\widehat{k},x \in R. If d→\overrightarrow{d} is the unit vector in the direction of b→+c→\overrightarrow{b}+\overrightarrow{c} such that a→⋅d→=1\overrightarrow{a}\cdot\overrightarrow{d}= 1, then (a→×b→)⋅c→(\overrightarrow{a}\times\overrightarrow{b}) \cdot\overrightarrow{c} is equal to

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  3. Question 3

    8 Apr 2024

    152
    Let  a→=i^+2j^+3k^, b→=2i^+3j^−5k^\ \overrightarrow{a}=\widehat{i}+ 2\widehat{j}+ 3\widehat{k},\ \overrightarrow{b}= 2\widehat{i}+ 3\widehat{j}- 5\widehat{k} and c→=3i^−j^+λk^\overrightarrow{c}= 3\widehat{i}-\widehat{j}+ \lambda\widehat{k} be three vectors. Let r→\overrightarrow{r} be a unit vector along b→+c→\overrightarrow{b}+\overrightarrow{c}. If r→⋅a→=3\overrightarrow{r}\cdot\overrightarrow{a}= 3, then 3λ3\lambda is equal to:

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  4. Question 4

    22 Jan 2026 S1

    51
    Let AB→=2i^+4j^−5k^\overrightarrow{AB}= 2\widehat{i}+ 4\widehat{j}- 5\widehat{k} and AD→=i^+2j^+λk^,λ∈R\overrightarrow{AD}=\widehat{i}+ 2\widehat{j}+ \lambda\widehat{k},\lambda \in R. Let the projection of the vector v→=i^+j^+k^\overrightarrow{v}=\widehat{i}+\widehat{j}+\widehat{k} on the diagonal AC→\overrightarrow{AC} of the parallelogram ABCD be of length one unit. If α,β\alpha,\beta, where α>β\alpha > \beta, be the roots of the equation λ2x2−6λx+5=0\lambda^{2}x^{2}- 6\lambda x + 5 = 0, then 2α−β2\alpha - \beta is equal to

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