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Expressions in the reciprocal vectors p = (b x c)/[a b c], q, r

This type of question, asked 4 times: 3 May 2023 S1, 4 May 2023 S2, 11 May 2023 S1, 20 Apr 2026 S2. All 4 questions below, oldest first.

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  1. Question 1

    3 May 2023 S1

    148
    Let a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} be three non-coplanar vectors and p⃗,q⃗,r⃗\vec{p},\vec{q},\vec{r} defined by p⃗=b⃗×c⃗[a⃗ b⃗ c⃗], q⃗=c⃗×a⃗[a⃗ b⃗ c⃗], r⃗=a⃗×b⃗[a⃗ b⃗ c⃗]\vec{p}=\frac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]},\ \vec{q}=\frac{\vec{c}\times\vec{a}}{[\vec{a}\ \vec{b}\ \vec{c}]},\ \vec{r}=\frac{\vec{a}\times\vec{b}}{[\vec{a}\ \vec{b}\ \vec{c}]}, then the value of (a⃗+b⃗)⋅p⃗+(b⃗+c⃗)⋅q⃗+(c⃗+a⃗)⋅r⃗(\vec{a}+\vec{b})\cdot\vec{p}+(\vec{b}+\vec{c})\cdot\vec{q}+(\vec{c}+\vec{a})\cdot\vec{r} is equal to

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  2. Question 2

    4 May 2023 S2

    106
    If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are non-coplanar vectors and p⃗=b⃗×c⃗[a⃗b⃗c⃗],q⃗=c⃗×a⃗[a⃗b⃗c⃗],r⃗=a⃗×b⃗[a⃗b⃗c⃗]\vec{p}=\frac{\vec{b}\times\vec{c}}{[\vec{a}\vec{b}\vec{c}]}, \vec{q}=\frac{\vec{c}\times\vec{a}}{[\vec{a}\vec{b}\vec{c}]}, \vec{r}=\frac{\vec{a}\times\vec{b}}{[\vec{a}\vec{b}\vec{c}]}, then 2a⃗⋅p⃗+b⃗⋅q⃗+c⃗⋅r⃗=2\vec{a}\cdot\vec{p}+\vec{b}\cdot\vec{q}+\vec{c}\cdot\vec{r}=

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  3. Question 3

    11 May 2023 S1

    132
    If x⃗=b⃗×c⃗[a⃗b⃗c⃗]\vec{x} = \frac{\vec{b}\times\vec{c}}{[\vec{a}\vec{b}\vec{c}]}, y⃗=c⃗×a⃗[a⃗b⃗c⃗]\vec{y} = \frac{\vec{c}\times\vec{a}}{[\vec{a}\vec{b}\vec{c}]} and z⃗=a⃗×b⃗[a⃗b⃗c⃗]\vec{z} = \frac{\vec{a}\times\vec{b}}{[\vec{a}\vec{b}\vec{c}]} where a, b, c are non-coplanar vectors, then value of x⃗⋅(a⃗+b⃗)+y⃗⋅(b⃗+c⃗)+z⃗⋅(c⃗+a⃗)\vec{x}\cdot(\vec{a}+\vec{b})+\vec{y}\cdot(\vec{b}+\vec{c})+\vec{z}\cdot(\vec{c}+\vec{a}) is

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  4. Question 4

    20 Apr 2026 S2

    90
    If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are three non-coplanar vectors and p⃗,q⃗,r⃗\vec{p}, \vec{q}, \vec{r} are defined as p⃗=b⃗×c⃗[a⃗ b⃗ c⃗],q⃗=c⃗×a⃗[a⃗ b⃗ c⃗],r⃗=a⃗×b⃗[a⃗ b⃗ c⃗]\vec{p} = \dfrac{\vec{b} \times \vec{c}}{[\vec{a}\,\vec{b}\,\vec{c}]}, \vec{q} = \dfrac{\vec{c} \times \vec{a}}{[\vec{a}\,\vec{b}\,\vec{c}]}, \vec{r} = \dfrac{\vec{a} \times \vec{b}}{[\vec{a}\,\vec{b}\,\vec{c}]}, then [(a⃗+b⃗)⋅p⃗+(b⃗+c⃗)⋅q⃗+(c⃗+a⃗)⋅r⃗][(\vec{a} + \vec{b})\cdot\vec{p} + (\vec{b} + \vec{c})\cdot\vec{q} + (\vec{c} + \vec{a})\cdot\vec{r}] is equal to

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