15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T4bd,

Given $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar vectors.

The three vectors are $\mathbf{u}=\mathbf{a}+2 \mathbf{b}+3 \mathbf{c}, \mathbf{v}=\lambda \mathbf{b}+4 \mathbf{c}$ and $\mathbf{w}=(2 \lambda-1) \mathbf{c}$

For these vectors to be non-coplanar, their scalar triple product must be non-zero.

$[\mathbf{u} \mathbf{v} \mathbf{w}] \neq \mathbf{0}$

This can be expressed as the determinant of the coefficients of $\mathbf{a}, \mathbf{b}, \mathbf{c}$

$$ \left|\begin{array}{ccc} 1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & 2 \lambda-1 \end{array}\right|[\mathbf{a} \mathbf{b} \mathbf{c}] \neq 0 $$

Since, $\mathbf{a , b , c}$ are non-coplanar, $[\mathbf{a} \mathbf{b} \mathbf{c}] \neq 0$

Therefore, we require the determinant to be non-zero.

$$ \begin{array}{r} 1 \cdot(\lambda(2 \lambda-1)-4 \cdot 0) \neq 0 \\ \lambda(2 \lambda-1) \neq 0 \end{array} $$

This implies $\lambda \neq 0$ and $2 \lambda-1 \neq 0$

$$ \Rightarrow \quad \lambda \neq \frac{1}{2} $$

Thus, the vectors are non-coplanar for all real values of $\lambda$ except 0 and $\frac{1}{2}$

This corresponds to "all except two values of $\lambda$ ".

Vector Algebra

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