Way Too Many Crossproducts!

Geometry Level 2

Suppose p ⃗ , q ⃗ , and r ⃗ \vec{p} , \vec{q}, \text{ and } \vec{r} are three mutually perpendicular unit vectors.

Vector u ⃗ \vec{u} satisfies the equation p ⃗ × ( ( u ⃗ − q ⃗ ) × p ⃗ ) + q ⃗ × ( ( u ⃗ − r ⃗ ) × q ⃗ ) + r ⃗ × ( ( u ⃗ − p ⃗ ) × r ⃗ ) = 0 \vec{p}\times((\vec{u} - \vec{q})\times\vec{p}) \hspace{.15cm} + \hspace{.15cm} \vec{q}\times((\vec{u} - \vec{r})\times\vec{q}) \hspace{.15cm} + \hspace{.15cm}\vec{r}\times((\vec{u} - \vec{p})\times\vec{r}) = 0

What is u ⃗ \vec{u} in terms of p ⃗ , q ⃗ , and r ⃗ \vec{p} , \vec{q}, \text{ and } \vec{r} ?

1 2 ( p ⃗ + q ⃗ + r ⃗ ) \frac{1}{2}(\vec{p} + \vec{q} + \vec{r}) 1 3 ( p ⃗ + q ⃗ + r ⃗ ) \frac{1}{3}(\vec{p} + \vec{q} + \vec{r}) 1 3 ( 2 p ⃗ + q ⃗ − r ⃗ ) \frac{1}{3}(2\vec{p} + \vec{q} - \vec{r}) 1 2 ( p ⃗ + q ⃗ − 2 r ⃗ ) \frac{1}{2}(\vec{p} + \vec{q} - 2\vec{r})

Harshit Damani by Harshit Damani , 25, India

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1 solution

Sandeep Bhardwaj
Oct 3, 2015

Using the notion of the vector triple product, we can say that \[\begin{array}{} \vec{p} \times \left( (\vec{u} -\vec{q}) \times \vec{p} \right) & =(\vec{p} \cdot \vec{p}) (\vec{u}-\vec{q}) -\left( \vec{p} \cdot ( \vec{u}-\vec{q} ) \right)\vec{p} \\ & =| \vec{p} |^2 (\vec{u}-\vec{q})-\left( \vec{p} \cdot \vec{u} -\vec{p} \cdot \vec{q} \right) \vec{p} \\ & =| \vec{p} |^2 (\vec{u}-\vec{q})-\left( \vec{p} \cdot \vec{u} \right) \vec{p} \quad \quad [\because \vec{p} \cdot \vec{q} =0] \end{array} \] Similarly we get q ⃗ × ( ( u ⃗ − r ⃗ ) × q ⃗ ) = ∣ q ⃗ ∣ 2 ( u ⃗ − r ⃗ ) − ( q ⃗ ⋅ u ⃗ ) q ⃗ r ⃗ × ( ( u ⃗ − p ⃗ ) × r ⃗ ) = ∣ r ⃗ ∣ 2 ( u ⃗ − p ⃗ ) − ( r ⃗ ⋅ u ⃗ ) r ⃗ \vec{q}\times((\vec{u} - \vec{r})\times\vec{q}) = | \vec{q} |^2 (\vec{u}-\vec{r})-\left( \vec{q} \cdot \vec{u} \right) \vec{q} \\ \vec{r}\times((\vec{u} - \vec{p})\times\vec{r}) = | \vec{r} |^2 (\vec{u}-\vec{p})-\left( \vec{r} \cdot \vec{u} \right) \vec{r}

Substituting the above values, our equation now reads

∣ p ⃗ ∣ 2 ( u ⃗ − q ⃗ ) − ( p ⃗ ⋅ u ⃗ ) p ⃗ + ∣ q ⃗ ∣ 2 ( u ⃗ − r ⃗ ) − ( q ⃗ ⋅ u ⃗ ) q ⃗ + ∣ r ⃗ ∣ 2 ( u ⃗ − p ⃗ ) − ( r ⃗ ⋅ u ⃗ ) r ⃗ = 0 | \vec{p} |^2 (\vec{u}-\vec{q})-\left( \vec{p} \cdot \vec{u} \right) \vec{p}+| \vec{q} |^2 (\vec{u}-\vec{r})-\left( \vec{q} \cdot \vec{u} \right) \vec{q}+| \vec{r} |^2 (\vec{u}-\vec{p})-\left( \vec{r} \cdot \vec{u} \right) \vec{r}=0 Since we are given that ∣ p ⃗ ∣ 2 = ∣ q ⃗ ∣ 2 = ∣ r ⃗ ∣ 2 = 1 |\vec{p}|^2=|\vec{q}|^2=|\vec{r}|^2=1 , our equation now becomes 3 u ⃗ − p ⃗ − q ⃗ − r ⃗ = ( p ⃗ ⋅ u ⃗ ) p ⃗ + ( q ⃗ ⋅ u ⃗ ) q ⃗ + ( r ⃗ ⋅ u ⃗ ) r ⃗ − − − ( 1 ) 3\vec{u} -\vec{p}-\vec{q}-\vec{r}=\left( \vec{p} \cdot \vec{u} \right) \vec{p}+\left( \vec{q} \cdot \vec{u} \right) \vec{q}+\left( \vec{r} \cdot \vec{u} \right) \vec{r} \qquad ---(1) As p ⃗ , q ⃗ , r ⃗ \vec{p}, \vec{q}, \vec{r} are mutually perpendicular to each other, we can express u ⃗ = λ 1 p ⃗ + λ 2 q ⃗ + λ 3 r ⃗ \vec{u}=\lambda_1 \vec{p}+\lambda_2 \vec{q}+\lambda_3 \vec{r} , as a linear combination of p ⃗ , q ⃗ \vec{p}, \vec{q} and r ⃗ \vec{r} . Taking its scalar product with p ⃗ \vec{p} , we get \[ \begin{array}{} \vec{u} \cdot \vec{p} & =\lambda_1 (\vec{p} \cdot \vec{p})+\lambda_2 (\vec{p} \cdot \vec{q} )+\lambda_3 (\vec{p} \cdot \vec{r}) \\ &=\lambda_1 |\vec{p}|^2+0+0=\lambda_1 \end{array} \]

Hence we obtained λ 1 = u ⃗ ⋅ p ⃗ \lambda_1=\vec{u} \cdot \vec{p} . In the same manner, we get λ 2 = q ⃗ ⋅ u ⃗ \lambda_2=\vec{q} \cdot \vec{u} and λ 3 = r ⃗ ⋅ u ⃗ \lambda_3=\vec{r}\cdot \vec{u} . Now with the expression of linear expression, we get u ⃗ = ( p ⃗ ⋅ u ⃗ ) p ⃗ + ( q ⃗ ⋅ u ⃗ ) q ⃗ + ( r ⃗ ⋅ u ⃗ ) r ⃗ \vec{u}= ( \vec{p} \cdot \vec{u} ) \vec{p}+( \vec{q} \cdot \vec{u} ) \vec{q}+( \vec{r} \cdot \vec{u} ) \vec{r} . Substituting this value into the equation (1), we have 3 u ⃗ − p ⃗ − q ⃗ − r ⃗ = u ⃗ ⇒ 2 u ⃗ = p ⃗ + q ⃗ + r ⃗ 3\vec{u}-\vec{p}-\vec{q}-\vec{r}=\vec{u} \\ \Rightarrow 2\vec{u}=\vec{p}+\vec{q}+\vec{r}

Hence we reached to the conclusion that u ⃗ = p ⃗ + q ⃗ + r ⃗ 2 \boxed{ \vec{u}=\dfrac{\vec{p}+\vec{q}+\vec{r}}{2}} .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Usually for such questions, it generally helps to evaluate the direction in terms of p , q , r p, q, r directly. For example, take the dot product with respect to p p , and then use the fact that a ⋅ ( b × c ) = b ⋅ ( c × a ) = c ⋅ ( a × b ) a \cdot (b \times c) = b \cdot ( c \times a) = c \cdot ( a \times b) . The first term is 0, the second term is p ⋅ q × ( ( u − r ) × q ) = ( ( u − r ) × q ) ⋅ ( p × q ) = ( ( u − r ) × q ) ⋅ r = r ⋅ ( ( u − r ) × q ) = ( u − r ) ⋅ ( q × r ) = ( u − r ) ⋅ p p \cdot q \times ( (u-r) \times q) = ((u-r) \times q ) \cdot ( p \times q ) = ( (u-r) \times q) \cdot r = r \cdot ( (u-r) \times q ) = (u-r) \cdot ( q \times r ) = ( u-r) \cdot p . The third term is p ⋅ r × ( ( u − p ) × r ) = ( ( u − p ) × r ) ⋅ ( p × r ) = ( ( u − p ) × r ) ⋅ ( − q ) = − q ⋅ ( ( u − p ) × r ) = − ( u − p ) ⋅ ( r × q ) = ( u − p ) p ˙ p \cdot r \times ( (u - p ) \times r ) =( (u-p) \times r ) \cdot ( p \times r) = ( (u-p) \times r ) \cdot ( - q ) = -q \cdot ( (u-p) \times r ) = - (u-p) \cdot ( r \times q ) = (u-p) \dot p .

Thus, ( 2 u − r − p ) ⋅ p = 0 (2u - r- p ) \cdot p = 0 which implies that the direction of u u in terms of p p is 1 2 \frac{1}{2} .

Excellent solution. Evenly thought out

Chaitanya Jha - 2 weeks, 1 day ago

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