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Question: If $\vec{a}$, $\vec{b}$, $\vec{c}$ are three non-zero, non-coplanar vectors and $\vec{b_1} = \vec{b}...

If a⃗\vec{a}, b⃗\vec{b}, c⃗\vec{c} are three non-zero, non-coplanar vectors and b1⃗=b⃗−b⃗.a⃗∣a⃗∣2a⃗\vec{b_1} = \vec{b} - \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}\vec{a}, b2⃗=b⃗+b⃗.a⃗∣a⃗∣2a⃗\vec{b_2} = \vec{b} + \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}\vec{a} and

c1⃗=c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b⃗∣b⃗∣2b⃗\vec{c_1} = \vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b}}{|\vec{b}|^2}\vec{b}, c2⃗=c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b1⃗∣b1⃗∣2b1⃗\vec{c_2} = \vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}, c3⃗=c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b2⃗∣b2⃗∣2b2⃗\vec{c_3} = \vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b_2}}{|\vec{b_2}|^2}\vec{b_2}, c4⃗=a⃗−c⃗.a⃗∣a⃗∣2a⃗\vec{c_4} = \vec{a} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a}

Then which of the following is a set of mutually orthogonal vectors?

A

{a⃗,b1⃗,c1⃗\vec{a}, \vec{b_1}, \vec{c_1}}

B

{a⃗,b1⃗,c2⃗\vec{a}, \vec{b_1}, \vec{c_2}}

C

{a⃗,b2⃗,c3⃗\vec{a}, \vec{b_2}, \vec{c_3}}

D

{a⃗,b2⃗,c4⃗\vec{a}, \vec{b_2}, \vec{c_4}}

Answer

{a⃗,b1⃗,c2⃗\vec{a}, \vec{b_1}, \vec{c_2}}

Explanation

Solution

The problem asks us to identify a set of mutually orthogonal vectors from the given options. We are given three non-zero, non-coplanar vectors a⃗\vec{a}, b⃗\vec{b}, c⃗\vec{c} and several derived vectors.

Let's analyze the definitions of the derived vectors:

  1. b1⃗=b⃗−b⃗.a⃗∣a⃗∣2a⃗\vec{b_1} = \vec{b} - \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}\vec{a}

    This expression represents the component of b⃗\vec{b} that is orthogonal to a⃗\vec{a}. This is a standard step in the Gram-Schmidt orthogonalization process. To verify orthogonality: b1⃗.a⃗=(b⃗−b⃗.a⃗∣a⃗∣2a⃗).a⃗=b⃗.a⃗−b⃗.a⃗∣a⃗∣2(a⃗.a⃗)=b⃗.a⃗−b⃗.a⃗=0\vec{b_1} . \vec{a} = \left(\vec{b} - \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}\vec{a}\right) . \vec{a} = \vec{b}.\vec{a} - \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}(\vec{a}.\vec{a}) = \vec{b}.\vec{a} - \vec{b}.\vec{a} = 0. So, a⃗\vec{a} and b1⃗\vec{b_1} are orthogonal.

  2. b2⃗=b⃗+b⃗.a⃗∣a⃗∣2a⃗\vec{b_2} = \vec{b} + \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}\vec{a}

    Let's check the dot product with a⃗\vec{a}: b2⃗.a⃗=(b⃗+b⃗.a⃗∣a⃗∣2a⃗).a⃗=b⃗.a⃗+b⃗.a⃗∣a⃗∣2(a⃗.a⃗)=b⃗.a⃗+b⃗.a⃗=2(b⃗.a⃗)\vec{b_2} . \vec{a} = \left(\vec{b} + \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}\vec{a}\right) . \vec{a} = \vec{b}.\vec{a} + \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}(\vec{a}.\vec{a}) = \vec{b}.\vec{a} + \vec{b}.\vec{a} = 2(\vec{b}.\vec{a}). Since a⃗\vec{a} and b⃗\vec{b} are general non-zero vectors, b⃗.a⃗\vec{b}.\vec{a} is generally not zero. Therefore, a⃗\vec{a} and b2⃗\vec{b_2} are generally not orthogonal. This immediately rules out options (C) and (D).

Now let's examine the remaining options, (A) and (B). Both include a⃗\vec{a} and b1⃗\vec{b_1}, which we've established are orthogonal. We need to check the third vector in each set.

Option (A): {a⃗,b1⃗,c1⃗}\{\vec{a}, \vec{b_1}, \vec{c_1}\} c1⃗=c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b⃗∣b⃗∣2b⃗\vec{c_1} = \vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b}}{|\vec{b}|^2}\vec{b} Let's check c1⃗.a⃗\vec{c_1} . \vec{a}: c1⃗.a⃗=(c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b⃗∣b⃗∣2b⃗).a⃗\vec{c_1} . \vec{a} = \left(\vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b}}{|\vec{b}|^2}\vec{b}\right) . \vec{a} =c⃗.a⃗−c⃗.a⃗∣a⃗∣2(a⃗.a⃗)−c⃗.b⃗∣b⃗∣2(b⃗.a⃗)= \vec{c}.\vec{a} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}(\vec{a}.\vec{a}) - \frac{\vec{c}.\vec{b}}{|\vec{b}|^2}(\vec{b}.\vec{a}) =c⃗.a⃗−c⃗.a⃗−(c⃗.b⃗)(b⃗.a⃗)∣b⃗∣2= \vec{c}.\vec{a} - \vec{c}.\vec{a} - \frac{(\vec{c}.\vec{b})(\vec{b}.\vec{a})}{|\vec{b}|^2} =−(c⃗.b⃗)(b⃗.a⃗)∣b⃗∣2= - \frac{(\vec{c}.\vec{b})(\vec{b}.\vec{a})}{|\vec{b}|^2}. For this to be zero, either c⃗.b⃗=0\vec{c}.\vec{b}=0 or b⃗.a⃗=0\vec{b}.\vec{a}=0. These conditions are not generally true for arbitrary non-zero, non-coplanar vectors. Thus, option (A) is generally not a set of mutually orthogonal vectors.

Option (B): {a⃗,b1⃗,c2⃗}\{\vec{a}, \vec{b_1}, \vec{c_2}\} We already know a⃗⊥b1⃗\vec{a} \perp \vec{b_1}. Now let's check c2⃗\vec{c_2}: c2⃗=c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b1⃗∣b1⃗∣2b1⃗\vec{c_2} = \vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}\vec{b_1} This expression for c2⃗\vec{c_2} is precisely the Gram-Schmidt orthogonalization step to make c⃗\vec{c} orthogonal to both a⃗\vec{a} and b1⃗\vec{b_1}. Let's verify:

  1. c2⃗.a⃗\vec{c_2} . \vec{a}: c2⃗.a⃗=(c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b1⃗∣b1⃗∣2b1⃗).a⃗\vec{c_2} . \vec{a} = \left(\vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}\right) . \vec{a} =c⃗.a⃗−c⃗.a⃗∣a⃗∣2(a⃗.a⃗)−c⃗.b1⃗∣b1⃗∣2(b1⃗.a⃗)= \vec{c}.\vec{a} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}(\vec{a}.\vec{a}) - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}(\vec{b_1}.\vec{a}) Since b1⃗.a⃗=0\vec{b_1}.\vec{a} = 0: =c⃗.a⃗−c⃗.a⃗−c⃗.b1⃗∣b1⃗∣2(0)=0= \vec{c}.\vec{a} - \vec{c}.\vec{a} - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}(0) = 0. So, a⃗⊥c2⃗\vec{a} \perp \vec{c_2}.

  2. c2⃗.b1⃗\vec{c_2} . \vec{b_1}: c2⃗.b1⃗=(c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b1⃗∣b1⃗∣2b1⃗).b1⃗\vec{c_2} . \vec{b_1} = \left(\vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}\right) . \vec{b_1} =c⃗.b1⃗−c⃗.a⃗∣a⃗∣2(a⃗.b1⃗)−c⃗.b1⃗∣b1⃗∣2(b1⃗.b1⃗)= \vec{c}.\vec{b_1} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}(\vec{a}.\vec{b_1}) - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}(\vec{b_1}.\vec{b_1}) Since a⃗.b1⃗=0\vec{a}.\vec{b_1} = 0: =c⃗.b1⃗−c⃗.a⃗∣a⃗∣2(0)−c⃗.b1⃗∣b1⃗∣2∣b1⃗∣2= \vec{c}.\vec{b_1} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}(0) - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}|\vec{b_1}|^2 =c⃗.b1⃗−c⃗.b1⃗=0= \vec{c}.\vec{b_1} - \vec{c}.\vec{b_1} = 0. So, b1⃗⊥c2⃗\vec{b_1} \perp \vec{c_2}.

Since a⃗⊥b1⃗\vec{a} \perp \vec{b_1}, a⃗⊥c2⃗\vec{a} \perp \vec{c_2}, and b1⃗⊥c2⃗\vec{b_1} \perp \vec{c_2}, the set {a⃗,b1⃗,c2⃗}\{\vec{a}, \vec{b_1}, \vec{c_2}\} is a set of mutually orthogonal vectors. This is the result of applying the Gram-Schmidt orthogonalization process to the vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}. The condition that a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are non-zero and non-coplanar ensures that b1⃗\vec{b_1} and c2⃗\vec{c_2} are also non-zero vectors.

The final answer is B\boxed{\text{B}}.

Explanation of the solution:

The problem asks for a set of mutually orthogonal vectors. This means the dot product of any two distinct vectors in the set must be zero.

  1. Analyze b1⃗\vec{b_1} and b2⃗\vec{b_2} with respect to a⃗\vec{a}:

    • b1⃗=b⃗−b⃗.a⃗∣a⃗∣2a⃗\vec{b_1} = \vec{b} - \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}\vec{a}. Calculating b1⃗.a⃗=(b⃗−b⃗.a⃗∣a⃗∣2a⃗).a⃗=b⃗.a⃗−b⃗.a⃗∣a⃗∣2∣a⃗∣2=b⃗.a⃗−b⃗.a⃗=0\vec{b_1}.\vec{a} = (\vec{b} - \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}\vec{a}).\vec{a} = \vec{b}.\vec{a} - \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}|\vec{a}|^2 = \vec{b}.\vec{a} - \vec{b}.\vec{a} = 0. Thus, a⃗⊥b1⃗\vec{a} \perp \vec{b_1}.
    • b2⃗=b⃗+b⃗.a⃗∣a⃗∣2a⃗\vec{b_2} = \vec{b} + \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}\vec{a}. Calculating b2⃗.a⃗=(b⃗+b⃗.a⃗∣a⃗∣2a⃗).a⃗=b⃗.a⃗+b⃗.a⃗∣a⃗∣2∣a⃗∣2=b⃗.a⃗+b⃗.a⃗=2(b⃗.a⃗)\vec{b_2}.\vec{a} = (\vec{b} + \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}\vec{a}).\vec{a} = \vec{b}.\vec{a} + \frac{\vec{b}.\vec{a}}{|\vec{a}|^2}|\vec{a}|^2 = \vec{b}.\vec{a} + \vec{b}.\vec{a} = 2(\vec{b}.\vec{a}). This is generally not zero. Therefore, options (C) and (D) are incorrect as they include a⃗\vec{a} and b2⃗\vec{b_2}.

  2. Analyze c1⃗\vec{c_1} with respect to a⃗\vec{a} (for option A):

    • c1⃗=c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b⃗∣b⃗∣2b⃗\vec{c_1} = \vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b}}{|\vec{b}|^2}\vec{b}. Calculating c1⃗.a⃗=(c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b⃗∣b⃗∣2b⃗).a⃗=c⃗.a⃗−c⃗.a⃗−c⃗.b⃗∣b⃗∣2(b⃗.a⃗)=−(c⃗.b⃗)(b⃗.a⃗)∣b⃗∣2\vec{c_1}.\vec{a} = (\vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b}}{|\vec{b}|^2}\vec{b}).\vec{a} = \vec{c}.\vec{a} - \vec{c}.\vec{a} - \frac{\vec{c}.\vec{b}}{|\vec{b}|^2}(\vec{b}.\vec{a}) = -\frac{(\vec{c}.\vec{b})(\vec{b}.\vec{a})}{|\vec{b}|^2}. This is generally not zero. Therefore, option (A) is incorrect.

  3. Analyze c2⃗\vec{c_2} with respect to a⃗\vec{a} and b1⃗\vec{b_1} (for option B):

    • c2⃗=c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b1⃗∣b1⃗∣2b1⃗\vec{c_2} = \vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}.
    • c2⃗.a⃗=(c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b1⃗∣b1⃗∣2b1⃗).a⃗=c⃗.a⃗−c⃗.a⃗−c⃗.b1⃗∣b1⃗∣2(b1⃗.a⃗)\vec{c_2}.\vec{a} = (\vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}).\vec{a} = \vec{c}.\vec{a} - \vec{c}.\vec{a} - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}(\vec{b_1}.\vec{a}). Since b1⃗.a⃗=0\vec{b_1}.\vec{a}=0, this simplifies to 00. Thus, a⃗⊥c2⃗\vec{a} \perp \vec{c_2}.
    • c2⃗.b1⃗=(c⃗−c⃗.a⃗∣a⃗∣2a⃗−c⃗.b1⃗∣b1⃗∣2b1⃗).b1⃗=c⃗.b1⃗−c⃗.a⃗∣a⃗∣2(a⃗.b1⃗)−c⃗.b1⃗∣b1⃗∣2∣b1⃗∣2\vec{c_2}.\vec{b_1} = (\vec{c} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}).\vec{b_1} = \vec{c}.\vec{b_1} - \frac{\vec{c}.\vec{a}}{|\vec{a}|^2}(\vec{a}.\vec{b_1}) - \frac{\vec{c}.\vec{b_1}}{|\vec{b_1}|^2}|\vec{b_1}|^2. Since a⃗.b1⃗=0\vec{a}.\vec{b_1}=0, this simplifies to c⃗.b1⃗−c⃗.b1⃗=0\vec{c}.\vec{b_1} - \vec{c}.\vec{b_1} = 0. Thus, b1⃗⊥c2⃗\vec{b_1} \perp \vec{c_2}.

Since a⃗⊥b1⃗\vec{a} \perp \vec{b_1}, a⃗⊥c2⃗\vec{a} \perp \vec{c_2}, and b1⃗⊥c2⃗\vec{b_1} \perp \vec{c_2}, the set {a⃗,b1⃗,c2⃗}\{\vec{a}, \vec{b_1}, \vec{c_2}\} is a set of mutually orthogonal vectors. This construction is exactly the Gram-Schmidt orthogonalization process.

The final answer is (B)\boxed{\text{(B)}}