If {a}={{i}}-{{k}}, {b}=x {{i}}+{{j}}+(1-x) {{k}} and {c}=y {{i}}+x {{j}}+(1+x-y) {{k}} then {a} ·({b} × {c}) depends on

Correct Answer is : neither xxx nor yyy

Correct Answer is : neither xxx nor yyy

Test Index

MHTCET 10 May 2024 Shift 1 Paper

Question : 104 of 150

Marks: +1, -0

\quad \overline{a}=\hat{\mathrm{i}}-\hat{\mathrm{k}}, \overline{b}=x \hat{\mathrm{i}}+\hat{\mathrm{j}}+(1-x) \hat{\mathrm{k}} \quad

\overline{c}=y \hat{\mathrm{i}}+x \hat{\mathrm{j}}+(1+x-y) \hat{\mathrm{k}}

\overline{a} \cdot(\overline{b} \times \overline{c})

depends on

Solution:

Concept:

The scalar triple product

\overline{a} \cdot (\overline{b} \times \overline{c})

is the determinant of the matrix formed by the components of the three vectors.

Explanation:

Write the scalar triple product as a determinant:

[\overline{a}\,\overline{b}\,\overline{c}] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{vmatrix}

Apply the column operation

C_3 \to C_3 + C_1

= \begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1+x \end{vmatrix}

Expand along the first row:

= 1 \cdot \begin{vmatrix} 1 & 1 \\ x & 1+x \end{vmatrix} = 1(1+x-x) = 1

The value is always

1

, independent of both

x

and

y

Answer:

The expression depends on neither

x

nor

y

Correct option: C. neither

x

nor

y

Go to Question:

Prev Question

Next Question