If a, b, c are three non-zero vectors and n is a unit vector perpendicular to c such that a = b - n, ( 0) and b c = 12, then | c ( a b) | is equal to:

Given: a = b - n with 0, n is a unit vector perpendicular to c, and b c = 12. Find: | c ( a b) |. Using the vector triple product identity, a b = ( b - n) b = ( b b) - ( n b) = - n b since b b = 0. Now, c ( a b) = c (- n b) Apply x ( y z) = y( x z) - z( x y) to get c (- n b) = - n( c b) + b( c n) Since n is perpendicular to c, c n = 0. Also, c b = 12. Therefore, c ( a b) = -12 n Hence, | c ( a b) | = |-12| | n| = 12 because | n| = 1. Therefore, the correct option is C and the required value is 12. Given: a = b - n and n c. Find: | c ( a b) |. Observe that the b part vanishes in a b because ( b) b = 0 So only the - n part contributes, making a b = - n b. Then in c ( a b), the triple product reduces to a vector along n, with magnitude | c b| | n| = 12 1 = 12 because c n = 0. Therefore, the required value is 12, so the correct option is C. The solution appears unrelated to this question, so the answer is derived from the question data itself.

If a, b, c are three non-zero vectors and n is a unit vector perp… — Solution | JEE Mathematics

MCQMediumJEE 2023Cross Product

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non-zero vectors and $\hat{n}$ is a unit vector perpendicular to $\mathbf{c}$ such that $\mathbf{a} = \alpha \mathbf{b} - \hat{n}, \quad (\alpha \neq 0)$ and $\mathbf{b} \cdot \mathbf{c} = 12, \quad then \quad \left| \mathbf{c} \times (\mathbf{a} \times \mathbf{b}) \right|$ is equal to:

A
$15$
B
$9$
C
$12$
D
$6$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: $\mathbf{a} = \alpha \mathbf{b} - \hat{n}$ with $\alpha \neq 0$ , $\hat{n}$ is a unit vector perpendicular to $\mathbf{c}$ , and $\mathbf{b} \cdot \mathbf{c} = 12$ .

Find: $\left| \mathbf{c} \times (\mathbf{a} \times \mathbf{b}) \right|$ .

Using the vector triple product identity,

\mathbf{a} \times \mathbf{b} = (\alpha \mathbf{b} - \hat{n}) \times \mathbf{b} = \alpha (\mathbf{b} \times \mathbf{b}) - (\hat{n} \times \mathbf{b}) = -\hat{n} \times \mathbf{b}

since $\mathbf{b} \times \mathbf{b} = \mathbf{0}$ .

Now,

\mathbf{c} \times (\mathbf{a} \times \mathbf{b}) = \mathbf{c} \times (-\hat{n} \times \mathbf{b})

Apply

\mathbf{x} \times (\mathbf{y} \times \mathbf{z}) = \mathbf{y}(\mathbf{x} \cdot \mathbf{z}) - \mathbf{z}(\mathbf{x} \cdot \mathbf{y})

to get

\mathbf{c} \times (-\hat{n} \times \mathbf{b}) = -\hat{n}(\mathbf{c} \cdot \mathbf{b}) + \mathbf{b}(\mathbf{c} \cdot \hat{n})

Since $\hat{n}$ is perpendicular to $\mathbf{c}$ , $\mathbf{c} \cdot \hat{n} = 0$ . Also, $\mathbf{c} \cdot \mathbf{b} = 12$ . Therefore,

\mathbf{c} \times (\mathbf{a} \times \mathbf{b}) = -12\hat{n}

Hence,

\left| \mathbf{c} \times (\mathbf{a} \times \mathbf{b}) \right| = |-12|\,|\hat{n}| = 12

because $|\hat{n}| = 1$ .

Therefore, the correct option is C and the required value is $12$ .

Use the structure of the vector directly

Given: $\mathbf{a} = \alpha \mathbf{b} - \hat{n}$ and $\hat{n} \perp \mathbf{c}$ .

Find: $\left| \mathbf{c} \times (\mathbf{a} \times \mathbf{b}) \right|$ .

Observe that the $\alpha \mathbf{b}$ part vanishes in $\mathbf{a} \times \mathbf{b}$ because

(\alpha \mathbf{b}) \times \mathbf{b} = \mathbf{0}

So only the $-\hat{n}$ part contributes, making $\mathbf{a} \times \mathbf{b} = -\hat{n} \times \mathbf{b}$ .

Then in $\mathbf{c} \times (\mathbf{a} \times \mathbf{b})$ , the triple product reduces to a vector along $\hat{n}$ , with magnitude

|\mathbf{c} \cdot \mathbf{b}|\,|\hat{n}| = 12 \cdot 1 = 12

because $\mathbf{c} \cdot \hat{n} = 0$ .

Therefore, the required value is $12$ , so the correct option is C.

The solution appears unrelated to this question, so the answer is derived from the question data itself.

Common mistakes

Using $\mathbf{a} \times \mathbf{b} = \alpha (\mathbf{a} \times \mathbf{b}) - \hat{n} \times \mathbf{b}$ is wrong because $\mathbf{a}$ has already been replaced by $\alpha \mathbf{b} - \hat{n}$ . The correct expansion is $\mathbf{a} \times \mathbf{b} = (\alpha \mathbf{b} - \hat{n}) \times \mathbf{b}$ .
Forgetting that $\mathbf{b} \times \mathbf{b} = \mathbf{0}$ leads to an unnecessary extra term. Always eliminate the self-cross-product immediately.
Applying the vector triple product formula in the wrong order is a common error. For $\mathbf{c} \times (\hat{n} \times \mathbf{b})$ , use $\mathbf{x} \times (\mathbf{y} \times \mathbf{z}) = \mathbf{y}(\mathbf{x} \cdot \mathbf{z}) - \mathbf{z}(\mathbf{x} \cdot \mathbf{y})$ , not a cyclically guessed version.
Ignoring that $\hat{n}$ is a unit vector causes a magnitude mistake. After obtaining $-12\hat{n}$ , use $|\hat{n}| = 1$ , so the magnitude is $12$ , not an unknown multiple of $12$ .

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