Let 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: (D) neither x nor y. Here, [a b c]=|a ·(b × c)| Now, according to question, lll a b c ] = | 1 0 -1 x 1 1 - x y x 1+x-y | =1(1+x-y-x(1-x))-1 (x^2-y ) =1+x-y-x+x^2-x^2+y =1 = Independent of x and y .

Q39 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics

Let $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{k}}, \mathbf{b}=x \hat{\mathbf{i}}+\hat{\mathbf{j}}+(1-x) \hat{\mathbf{k}}$ and $\mathbf{c}=y \hat{\mathbf{i}}+x \hat{\mathbf{j}}+(1+x-y) \hat{\mathbf{k}}$. Then, $[\mathbf{a} \mathbf{b} \mathbf{~ c}]$ depends on

A
only $y$
B
only $x$
C
both $x$ and $y$
Correct answer
D
neither $x$ nor $y$

Solution

Here, $[a b c]=|a \cdot(b \times c)|$ Now, according to question, $\begin{aligned} {\left[\begin{array}{lll} a & b c \end{array}\right] } & =\left|\begin{array}{ccc} 1 & 0 & -1 \\ x & 1 & 1 - x \\ y & x & 1+x-y \end{array}\right| \\ & =1(1+x-y-x(1-x))-1\left(x^2-y\right) \\ & =1+x-y-x+x^2-x^2+y \\ & =1 \\ & =\text { Independent of } x \text { and } y . \end{aligned}$

Hence, the correct answer is option: 'D'.

Let $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{k}}, \mathbf{b}=x \hat{\mathbf{i}}+\hat{\mathbf{j}}+(1-x) \hat{\mathbf{k}}$ and $\mathbf{c}=y \hat{\mathbf{i}}+x \hat{\mathbf{j}}+(1+x-y) \hat{\mathbf{k}}$. Then, $[\mathbf{a} \mathbf{b} \mathbf{~ c}]$ depends on

Solution

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